\displaystyle \text{Sine and Cosine Formulae and their Application}

\displaystyle \text{Trigonometric Equations}

\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{In any }\triangle ABC,\ \sum a(\sin B-\sin C)=
\displaystyle \text{(a) }a^2+b^2+c^2\qquad\text{(b) }a^2\qquad\text{(c) }b^2\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \sum a(\sin B-\sin C)
\displaystyle =a(\sin B-\sin C)+b(\sin C-\sin A)+c(\sin A-\sin B).
\displaystyle \text{By the sine rule, let }a=k\sin A,\quad b=k\sin B,\quad c=k\sin C.
\displaystyle \therefore \sum a(\sin B-\sin C)
\displaystyle =k\{\sin A(\sin B-\sin C)+\sin B(\sin C-\sin A)
\displaystyle \quad+\sin C(\sin A-\sin B)\}
\displaystyle =k\{\sin A\sin B-\sin A\sin C+\sin B\sin C
\displaystyle \quad-\sin A\sin B+\sin A\sin C-\sin B\sin C\}
\displaystyle =0.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In a }\triangle ABC,\text{ if }a=2,\ \angle B=60^\circ\text{ and }\angle C=75^\circ,\text{ then }b=
\displaystyle \text{(a) }\sqrt3\qquad\text{(b) }\sqrt6\qquad\text{(c) }\sqrt9\qquad\text{(d) }1+\sqrt2
\displaystyle \text{Answer:}
\displaystyle A=180^\circ-(B+C)=180^\circ-(60^\circ+75^\circ)=45^\circ.
\displaystyle \text{By the sine rule, }\frac{a}{\sin A}=\frac{b}{\sin B}.
\displaystyle \therefore b=\frac{a\sin B}{\sin A}
\displaystyle =\frac{2\sin60^\circ}{\sin45^\circ}
\displaystyle =\frac{2\left(\frac{\sqrt3}{2}\right)}{\frac1{\sqrt2}}=\sqrt6.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the sides of a triangle are in the ratio }1:\sqrt3:2,\text{ then}
\displaystyle \text{the measure of its greatest angle is}
\displaystyle \text{(a) }\frac{\pi}{6}\qquad\text{(b) }\frac{\pi}{3}\qquad\text{(c) }\frac{\pi}{2}\qquad\text{(d) }\frac{2\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \text{The greatest side is }2.
\displaystyle 1^2+(\sqrt3)^2=1+3=4=2^2.
\displaystyle \therefore \text{The triangle is right-angled.}
\displaystyle \therefore \text{Its greatest angle}=\frac{\pi}{2}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In any }\triangle ABC,\ 2(bc\cos A+ca\cos B+ab\cos C)=
\displaystyle \text{(a) }abc\qquad\text{(b) }a+b+c\qquad\text{(c) }a^2+b^2+c^2
\displaystyle \text{(d) }\frac1{a^2}+\frac1{b^2}+\frac1{c^2}
\displaystyle \text{Answer:}
\displaystyle \text{By the cosine rule,}
\displaystyle 2bc\cos A=b^2+c^2-a^2,
\displaystyle 2ca\cos B=c^2+a^2-b^2,
\displaystyle 2ab\cos C=a^2+b^2-c^2.
\displaystyle \text{Adding,}
\displaystyle 2(bc\cos A+ca\cos B+ab\cos C)
\displaystyle =a^2+b^2+c^2.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In a triangle }ABC,\ a=4,\ b=3,\ \angle A=60^\circ\text{ then }c\text{ is a root}
\displaystyle \text{of the equation}
\displaystyle \text{(a) }c^2-3c-7=0\qquad\text{(b) }c^2+3c+7=0
\displaystyle \text{(c) }c^2-3c+7=0\qquad\text{(d) }c^2+3c-7=0
\displaystyle \text{Answer:}
\displaystyle \text{By the cosine rule, }a^2=b^2+c^2-2bc\cos A.
\displaystyle 4^2=3^2+c^2-2(3)c\cos60^\circ
\displaystyle 16=9+c^2-3c.
\displaystyle \therefore c^2-3c-7=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In a }\triangle ABC,\text{ if }(c+a+b)(a+b-c)=ab,\text{ then}
\displaystyle \text{the measure of angle }C\text{ is}
\displaystyle \text{(a) }\frac{\pi}{3}\qquad\text{(b) }\frac{\pi}{6}\qquad\text{(c) }\frac{2\pi}{3}\qquad\text{(d) }\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle (c+a+b)(a+b-c)=ab.
\displaystyle (a+b+c)(a+b-c)=ab
\displaystyle \Rightarrow (a+b)^2-c^2=ab.
\displaystyle \Rightarrow c^2=a^2+ab+b^2.
\displaystyle \text{By the cosine rule, }c^2=a^2+b^2-2ab\cos C.
\displaystyle \therefore a^2+b^2-2ab\cos C=a^2+ab+b^2.
\displaystyle \Rightarrow -2ab\cos C=ab
\displaystyle \Rightarrow \cos C=-\frac12.
\displaystyle \therefore C=\frac{2\pi}{3}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In any }\triangle ABC,\text{ the value of }2ac\sin\left(\frac{A-B+C}{2}\right)\text{ is}
\displaystyle \text{(a) }a^2+b^2-c^2\qquad\text{(b) }c^2+a^2-b^2
\displaystyle \text{(c) }b^2-c^2-a^2\qquad\text{(d) }c^2-a^2-b^2
\displaystyle \text{Answer:}
\displaystyle A+B+C=\pi.
\displaystyle \therefore A+C=\pi-B.
\displaystyle \therefore \frac{A-B+C}{2}=\frac{\pi-2B}{2}=\frac{\pi}{2}-B.
\displaystyle \therefore \sin\left(\frac{A-B+C}{2}\right)=\cos B.
\displaystyle \therefore 2ac\sin\left(\frac{A-B+C}{2}\right)=2ac\cos B.
\displaystyle \text{By the cosine rule, }b^2=a^2+c^2-2ac\cos B.
\displaystyle \therefore 2ac\cos B=a^2+c^2-b^2.
\displaystyle \therefore 2ac\sin\left(\frac{A-B+C}{2}\right)=a^2+c^2-b^2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In any }\triangle ABC,\ a(b\cos C-c\cos B)=
\displaystyle \text{(a) }a^2\qquad\text{(b) }b^2-c^2\qquad\text{(c) }0\qquad\text{(d) }b^2+c^2
\displaystyle \text{Answer:}
\displaystyle \text{By the cosine rule,}
\displaystyle \cos C=\frac{a^2+b^2-c^2}{2ab},
\displaystyle \cos B=\frac{a^2+c^2-b^2}{2ac}.
\displaystyle \therefore b\cos C=\frac{a^2+b^2-c^2}{2a}.
\displaystyle c\cos B=\frac{a^2+c^2-b^2}{2a}.
\displaystyle \therefore a(b\cos C-c\cos B)
\displaystyle =a\left\{\frac{a^2+b^2-c^2-a^2-c^2+b^2}{2a}\right\}
\displaystyle =a\left(\frac{2b^2-2c^2}{2a}\right)
\displaystyle =b^2-c^2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The smallest value of }x\text{ satisfying the equation}
\displaystyle \sqrt3(\cot x+\tan x)=4\text{ is}
\displaystyle \text{(a) }\frac{2\pi}{3}\qquad\text{(b) }\frac{\pi}{3}\qquad\text{(c) }\frac{\pi}{6}\qquad\text{(d) }\frac{\pi}{12}
\displaystyle \text{Answer:}
\displaystyle \sqrt3(\cot x+\tan x)=4
\displaystyle \therefore \sqrt3\left(\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right)=4.
\displaystyle \therefore \frac{\sqrt3}{\sin x\cos x}=4.
\displaystyle \therefore \sin2x=\frac{\sqrt3}{2}.
\displaystyle \therefore 2x=n\pi+(-1)^n\frac{\pi}{3},\qquad n\in Z.
\displaystyle \text{The smallest positive value of }2x\text{ is }\frac{\pi}{3}.
\displaystyle \therefore x=\frac{\pi}{6}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\cos x+\sqrt3\sin x=2,\text{ then }x=
\displaystyle \text{(a) }\frac{\pi}{3}\qquad\text{(b) }\frac{2\pi}{3}\qquad\text{(c) }\frac{4\pi}{3}\qquad\text{(d) }\frac{5\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \cos x+\sqrt3\sin x=2.
\displaystyle \therefore 2\left(\frac12\cos x+\frac{\sqrt3}{2}\sin x\right)=2.
\displaystyle \therefore 2\sin\left(x+\frac{\pi}{6}\right)=2.
\displaystyle \therefore \sin\left(x+\frac{\pi}{6}\right)=1.
\displaystyle \therefore x+\frac{\pi}{6}=2n\pi+\frac{\pi}{2},\qquad n\in Z.
\displaystyle \therefore x=2n\pi+\frac{\pi}{3}.
\displaystyle \therefore \text{From the given options, }x=\frac{\pi}{3}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\tan px-\tan qx=0,\text{ then the values of }x
\displaystyle \text{form a series in}
\displaystyle \text{(a) AP}\qquad\text{(b) GP}\qquad\text{(c) HP}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \tan px-\tan qx=0
\displaystyle \therefore \tan px=\tan qx.
\displaystyle \therefore px=n\pi+qx,\qquad n\in Z.
\displaystyle \therefore x=\frac{n\pi}{p-q},\qquad p\ne q.
\displaystyle \therefore x=0,\frac{\pi}{p-q},\frac{2\pi}{p-q},\ldots
\displaystyle \text{These values have a constant common difference.}
\displaystyle \therefore \text{the values of }x\text{ form an AP.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }a\text{ is any real number, the number of roots of}
\displaystyle \cot x-\tan x=a\text{ in the first quadrant is (are).}
\displaystyle \text{(a) }2\qquad\text{(b) }0\qquad\text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cot x-\tan x=\frac{\cos^2x-\sin^2x}{\sin x\cos x}
\displaystyle =\frac{\cos2x}{\frac12\sin2x}=2\cot2x.
\displaystyle \therefore 2\cot2x=a.
\displaystyle \therefore \cot2x=\frac{a}{2}.
\displaystyle \text{In the first quadrant, }0<x<\frac{\pi}{2},\text{ so }0<2x<\pi.
\displaystyle \cot2x\text{ takes every real value exactly once in }(0,\pi).
\displaystyle \therefore \text{there is exactly one root for every real }a.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The general solution of the equation}
\displaystyle 7\cos^2x+3\sin^2x=4\text{ is}
\displaystyle \text{(a) }x=2n\pi\pm\frac{\pi}{6},\ n\in Z
\displaystyle \text{(b) }x=2n\pi\pm\frac{2\pi}{3},\ n\in Z
\displaystyle \text{(c) }x=n\pi\pm\frac{\pi}{3},\ n\in Z
\displaystyle \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 7\cos^2x+3\sin^2x=4.
\displaystyle \therefore 7\cos^2x+3(1-\cos^2x)=4.
\displaystyle \therefore 4\cos^2x=1.
\displaystyle \therefore \cos^2x=\frac14=\cos^2\frac{\pi}{3}.
\displaystyle \text{Using }\cos^2x=\cos^2\alpha\Rightarrow x=n\pi\pm\alpha,
\displaystyle \therefore x=n\pi\pm\frac{\pi}{3},\qquad n\in Z.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A solution of the equation }\cos^2x+\sin x+1=0,
\displaystyle \text{lies in the interval}
\displaystyle \text{(a) }\left(-\frac{\pi}{4},\frac{\pi}{4}\right)\qquad \text{(b) }\left(\frac{\pi}{4},\frac{3\pi}{4}\right)
\displaystyle \text{(c) }\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\qquad \text{(d) }\left(\frac{5\pi}{4},\frac{7\pi}{4}\right)
\displaystyle \text{Answer:}
\displaystyle \cos^2x+\sin x+1=0.
\displaystyle \therefore 1-\sin^2x+\sin x+1=0.
\displaystyle \therefore \sin^2x-\sin x-2=0.
\displaystyle \therefore (\sin x-2)(\sin x+1)=0.
\displaystyle \therefore \sin x=-1,\quad\text{since }\sin x=2\text{ is not possible.}
\displaystyle \therefore x=\frac{3\pi}{2}\text{ is a solution.}
\displaystyle \frac{5\pi}{4}<\frac{3\pi}{2}<\frac{7\pi}{4}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The number of solution in }[0,\frac{\pi}{2}]\text{ of the}
\displaystyle \text{equation }\cos3x\tan5x=\sin7x\text{ is}
\displaystyle \text{(a) }5\qquad\text{(b) }7\qquad\text{(c) }6\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cos3x\tan5x=\sin7x,\qquad \cos5x\ne0.
\displaystyle \therefore \cos3x\sin5x=\sin7x\cos5x.
\displaystyle \therefore \sin8x+\sin2x=\sin12x+\sin2x.
\displaystyle \therefore \sin8x=\sin12x.
\displaystyle \therefore \sin8x-\sin12x=0.
\displaystyle \therefore -2\cos10x\sin2x=0.
\displaystyle \therefore \sin2x=0\quad\text{or}\quad\cos10x=0.
\displaystyle \sin2x=0\Rightarrow x=0,\frac{\pi}{2}\text{ in }[0,\frac{\pi}{2}].
\displaystyle x=\frac{\pi}{2}\text{ is rejected since }\tan5x\text{ is undefined.}
\displaystyle \therefore \sin2x=0\text{ gives one valid solution, }x=0.
\displaystyle \cos10x=0\Rightarrow x=\frac{(2n+1)\pi}{20}.
\displaystyle \text{In }[0,\frac{\pi}{2}],\quad x=\frac{\pi}{20},\frac{3\pi}{20},\frac{5\pi}{20},\frac{7\pi}{20},\frac{9\pi}{20}.
\displaystyle \therefore \cos10x=0\text{ gives }5\text{ valid solutions.}
\displaystyle \therefore \text{the total number of solutions is }1+5=6.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The general value of }x\text{ satisfying the equation}
\displaystyle \sqrt3\sin x+\cos x=\sqrt3\text{ is given by}
\displaystyle \text{(a) }x=n\pi+(-1)^n\frac{\pi}{4}+\frac{\pi}{3},\ n\in Z
\displaystyle \text{(b) }x=n\pi+(-1)^n\frac{\pi}{3}-\frac{\pi}{6},\ n\in Z
\displaystyle \text{(c) }x=n\pi\pm\frac{\pi}{6},\ n\in Z
\displaystyle \text{(d) }x=n\pi\pm\frac{\pi}{3},\ n\in Z
\displaystyle \text{Answer:}
\displaystyle \sqrt3\sin x+\cos x=\sqrt3.
\displaystyle \therefore 2\left(\frac{\sqrt3}{2}\sin x+\frac12\cos x\right)=\sqrt3.
\displaystyle \therefore 2\sin\left(x+\frac{\pi}{6}\right)=\sqrt3.
\displaystyle \therefore \sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt3}{2}=\sin\frac{\pi}{3}.
\displaystyle \text{Using }\sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha,
\displaystyle x+\frac{\pi}{6}=n\pi+(-1)^n\frac{\pi}{3},\qquad n\in Z.
\displaystyle \therefore x=n\pi+(-1)^n\frac{\pi}{3}-\frac{\pi}{6},\qquad n\in Z.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The smallest positive angle which satisfies the equation}
\displaystyle 2\sin^2x+\sqrt3\cos x+1=0\text{ is}
\displaystyle \text{(a) }\frac{5\pi}{6}\qquad\text{(b) }\frac{2\pi}{3}\qquad\text{(c) }\frac{\pi}{3}\qquad\text{(d) }\frac{\pi}{6}
\displaystyle \text{Answer:}
\displaystyle 2\sin^2x+\sqrt3\cos x+1=0.
\displaystyle \therefore 2(1-\cos^2x)+\sqrt3\cos x+1=0.
\displaystyle \therefore 2\cos^2x-\sqrt3\cos x-3=0.
\displaystyle \therefore (2\cos x+\sqrt3)(\cos x-\sqrt3)=0.
\displaystyle \therefore \cos x=-\frac{\sqrt3}{2}\quad\text{or}\quad\cos x=\sqrt3.
\displaystyle \cos x=\sqrt3\text{ is not possible since }-1\le\cos x\le1.
\displaystyle \therefore \cos x=-\frac{\sqrt3}{2}.
\displaystyle \text{The smallest positive value of }x\text{ is }\frac{5\pi}{6}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }4\sin^2x=1,\text{ then the values of }x\text{ are}
\displaystyle \text{(a) }2n\pi\pm\frac{\pi}{3},\ n\in Z\qquad \text{(b) }n\pi\pm\frac{\pi}{3},\ n\in Z
\displaystyle \text{(c) }n\pi\pm\frac{\pi}{6},\ n\in Z\qquad \text{(d) }2n\pi\pm\frac{\pi}{6},\ n\in Z
\displaystyle \text{Answer:}
\displaystyle 4\sin^2x=1.
\displaystyle \therefore \sin^2x=\frac14=\sin^2\frac{\pi}{6}.
\displaystyle \text{Using }\sin^2x=\sin^2\alpha\Rightarrow x=n\pi\pm\alpha,
\displaystyle \therefore x=n\pi\pm\frac{\pi}{6},\qquad n\in Z.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\cot x-\tan x=\sec x,\text{ then }x\text{ is equal to}
\displaystyle \text{(a) }2n\pi+\frac{3\pi}{2},\ n\in Z\qquad \text{(b) }n\pi+(-1)^n\frac{\pi}{6},\ n\in Z
\displaystyle \text{(c) }n\pi\pm\frac{\pi}{2},\ n\in Z\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cot x-\tan x=\sec x.
\displaystyle \therefore \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=\frac{1}{\cos x}.
\displaystyle \therefore \cos^2x-\sin^2x=\sin x.
\displaystyle \therefore \cos2x=\sin x.
\displaystyle \therefore 1-2\sin^2x=\sin x.
\displaystyle \therefore 2\sin^2x+\sin x-1=0.
\displaystyle \therefore (2\sin x-1)(\sin x+1)=0.
\displaystyle \therefore \sin x=\frac12\quad\text{or}\quad\sin x=-1.
\displaystyle \sin x=-1\Rightarrow \cos x=0,\text{ for which }\sec x\text{ is undefined.}
\displaystyle \therefore \sin x=\frac12.
\displaystyle \therefore x=n\pi+(-1)^n\frac{\pi}{6},\qquad n\in Z.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A value of }x\text{ satisfying }\cos x+\sqrt3\sin x=2\text{ is}
\displaystyle \text{(a) }\frac{5\pi}{3}\qquad\text{(b) }\frac{4\pi}{3}\qquad \text{(c) }\frac{2\pi}{3}\qquad\text{(d) }\frac{\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \cos x+\sqrt3\sin x=2.
\displaystyle \therefore 2\left(\frac12\cos x+\frac{\sqrt3}{2}\sin x\right)=2.
\displaystyle \therefore 2\sin\left(x+\frac{\pi}{6}\right)=2.
\displaystyle \therefore \sin\left(x+\frac{\pi}{6}\right)=1.
\displaystyle \therefore x+\frac{\pi}{6}=2n\pi+\frac{\pi}{2},\qquad n\in Z.
\displaystyle \therefore x=2n\pi+\frac{\pi}{3}.
\displaystyle \therefore \text{From the given options, }x=\frac{\pi}{3}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In }(0,\pi),\text{ the number of solutions of the equation}
\displaystyle \tan x+\tan2x+\tan3x=\tan x\tan2x\tan3x\text{ is}
\displaystyle \text{(a) }7\qquad\text{(b) }5\qquad\text{(c) }4\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \tan A+\tan B+\tan C-\tan A\tan B\tan C
\displaystyle =\frac{\sin(A+B+C)}{\cos A\cos B\cos C}.
\displaystyle \therefore \sin(x+2x+3x)=0.
\displaystyle \therefore \sin6x=0.
\displaystyle \therefore 6x=n\pi\Rightarrow x=\frac{n\pi}{6}.
\displaystyle \text{For }0<x<\pi,\quad x=\frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2}, \frac{2\pi}{3},\frac{5\pi}{6}.
\displaystyle x=\frac{\pi}{6},\frac{5\pi}{6}\text{ are rejected since }\tan3x\text{ is undefined.}
\displaystyle x=\frac{\pi}{2}\text{ is rejected since }\tan x\text{ is undefined.}
\displaystyle \therefore x=\frac{\pi}{3},\frac{2\pi}{3}\text{ are the only valid solutions.}
\displaystyle \therefore \text{the number of solutions is }2.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The number of values of }x\text{ in }[0,2\pi]\text{ that satisfy}
\displaystyle \text{the equation }\sin^2x-\cos x=\frac14\text{ is}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \sin^2x-\cos x=\frac14.
\displaystyle \therefore 1-\cos^2x-\cos x=\frac14.
\displaystyle \therefore 4\cos^2x+4\cos x-3=0.
\displaystyle \therefore (2\cos x-1)(2\cos x+3)=0.
\displaystyle \therefore \cos x=\frac12\quad\text{or}\quad\cos x=-\frac32.
\displaystyle \cos x=-\frac32\text{ is not possible since }-1\le\cos x\le1.
\displaystyle \therefore \cos x=\frac12.
\displaystyle \text{For }x\in[0,2\pi],\quad x=\frac{\pi}{3},\frac{5\pi}{3}.
\displaystyle \therefore \text{the number of values of }x\text{ is }2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }e^{\sin x}-e^{-\sin x}-4=0,\text{ then }x=
\displaystyle \text{(a) }0\qquad\text{(b) }\sin^{-1}\{\log_e(2-\sqrt5)\}\qquad \text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle e^{\sin x}-e^{-\sin x}-4=0.
\displaystyle \text{Let }y=e^{\sin x},\text{ where }y>0.
\displaystyle \therefore y-\frac1y-4=0.
\displaystyle \therefore y^2-4y-1=0.
\displaystyle \therefore y=2\pm\sqrt5.
\displaystyle \text{Since }y>0,\quad y=2+\sqrt5.
\displaystyle \therefore e^{\sin x}=2+\sqrt5.
\displaystyle \therefore \sin x=\log_e(2+\sqrt5).
\displaystyle \text{But }\log_e(2+\sqrt5)>1,\text{ whereas }-1\le\sin x\le1.
\displaystyle \therefore \text{there is no real value of }x.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The equation }3\cos x+4\sin x=6\text{ has .... solution}
\displaystyle \text{(a) finite}\qquad\text{(b) infinite}\qquad \text{(c) one}\qquad\text{(d) no}
\displaystyle \text{Answer:}
\displaystyle |3\cos x+4\sin x|\le\sqrt{3^2+4^2}=5.
\displaystyle \therefore -5\le3\cos x+4\sin x\le5.
\displaystyle \text{Since }6>5,\text{ the equation has no real solution.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }\sqrt3\cos x+\sin x=\sqrt2,\text{ then general value of }x\text{ is}
\displaystyle \text{(a) }n\pi+(-1)^n\frac{\pi}{4}+\frac{\pi}{6},\ n\in Z
\displaystyle \text{(b) }(-1)^n\frac{\pi}{4}-\frac{\pi}{3},\ n\in Z
\displaystyle \text{(c) }n\pi+\frac{\pi}{4}-\frac{\pi}{3},\ n\in Z
\displaystyle \text{(d) }n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{3},\ n\in Z
\displaystyle \text{Answer:}
\displaystyle \sqrt3\cos x+\sin x=\sqrt2.
\displaystyle \therefore 2\cos\left(x-\frac{\pi}{6}\right)=\sqrt2.
\displaystyle \therefore \cos\left(x-\frac{\pi}{6}\right)=\frac{1}{\sqrt2}=\cos\frac{\pi}{4}.
\displaystyle \therefore x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{4},\qquad n\in Z.
\displaystyle \therefore x=2n\pi+\frac{\pi}{6}\pm\frac{\pi}{4}.
\displaystyle \therefore x=2n\pi+\frac{5\pi}{12}\quad\text{or}\quad 2n\pi-\frac{\pi}{12}.
\displaystyle \text{These two families can be written together as}
\displaystyle x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{3},\qquad n\in Z.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{General solution of }\tan5x=\cot2x\text{ is}
\displaystyle \text{(a) }\frac{n\pi}{7}+\frac{\pi}{2},\ n\in Z\qquad \text{(b) }x=\frac{n\pi}{7}+\frac{\pi}{3},\ n\in Z
\displaystyle \text{(c) }x=\frac{n\pi}{7}+\frac{\pi}{14},\ n\in Z\qquad \text{(d) }x=\frac{n\pi}{7}-\frac{\pi}{14},\ n\in Z
\displaystyle \text{Answer:}
\displaystyle \tan5x=\cot2x.
\displaystyle \therefore \tan5x=\tan\left(\frac{\pi}{2}-2x\right).
\displaystyle \therefore 5x=n\pi+\frac{\pi}{2}-2x,\qquad n\in Z.
\displaystyle \therefore 7x=n\pi+\frac{\pi}{2}.
\displaystyle \therefore x=\frac{n\pi}{7}+\frac{\pi}{14},\qquad n\in Z.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{The solution of the equation }\cos^2x+\sin x+1=0
\displaystyle \text{lies in the interval}
\displaystyle \text{(a) }\left(-\frac{\pi}{4},\frac{\pi}{4}\right)\qquad \text{(b) }\left(\frac{\pi}{4},\frac{3\pi}{4}\right)
\displaystyle \text{(c) }\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\qquad \text{(d) }\left(\frac{5\pi}{4},\frac{7\pi}{4}\right)
\displaystyle \text{Answer:}
\displaystyle \cos^2x+\sin x+1=0.
\displaystyle \therefore 1-\sin^2x+\sin x+1=0.
\displaystyle \therefore \sin^2x-\sin x-2=0.
\displaystyle \therefore (\sin x-2)(\sin x+1)=0.
\displaystyle \therefore \sin x=2\quad\text{or}\quad\sin x=-1.
\displaystyle \sin x=2\text{ is not possible since }-1\le\sin x\le1.
\displaystyle \therefore \sin x=-1\Rightarrow x=\frac{3\pi}{2}+2n\pi.
\displaystyle \frac{5\pi}{4}<\frac{3\pi}{2}<\frac{7\pi}{4}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }\cos x=-\frac12\text{ and }0<x<2\pi,\text{ then the solutions are}
\displaystyle \text{(a) }x=\frac{\pi}{3},\frac{4\pi}{3}\qquad \text{(b) }x=\frac{2\pi}{3},\frac{4\pi}{3}
\displaystyle \text{(c) }x=\frac{2\pi}{3},\frac{7\pi}{6}\qquad \text{(d) }\theta=\frac{2\pi}{3},\frac{5\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \cos x=-\frac12=\cos\frac{2\pi}{3}.
\displaystyle \text{For }0<x<2\pi,\quad x=\frac{2\pi}{3},\frac{4\pi}{3}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{The number of values of }x\text{ in the interval }[0,5\pi]
\displaystyle \text{satisfying the equation }3\sin^2x-7\sin x+2=0\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }5\qquad\text{(c) }6\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle 3\sin^2x-7\sin x+2=0.
\displaystyle \therefore (3\sin x-1)(\sin x-2)=0.
\displaystyle \therefore \sin x=\frac13\quad\text{or}\quad\sin x=2.
\displaystyle \sin x=2\text{ is not possible since }-1\le\sin x\le1.
\displaystyle \therefore \sin x=\frac13.
\displaystyle \text{There are two solutions in each positive half-cycle of }\sin x.
\displaystyle \text{In }[0,5\pi],\text{ these occur in }(0,\pi),(2\pi,3\pi),(4\pi,5\pi).
\displaystyle \therefore \text{the number of solutions is }2+2+2=6.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Find the area of the triangle }\triangle ABC\text{ in which }a=1,\ b=2
\displaystyle \text{and }\angle C=60^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=\frac12ab\sin C.
\displaystyle =\frac12(1)(2)\sin60^\circ
\displaystyle =\frac{\sqrt3}{2}.
\displaystyle \therefore \text{Area of }\triangle ABC=\frac{\sqrt3}{2}\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In a }\triangle ABC,\text{ if }b=\sqrt3,\ c=1\text{ and }\angle A=30^\circ,\text{ find }a.
\displaystyle \text{Answer:}
\displaystyle \text{By the cosine rule, }a^2=b^2+c^2-2bc\cos A.
\displaystyle a^2=(\sqrt3)^2+1^2-2(\sqrt3)(1)\cos30^\circ
\displaystyle =3+1-2\sqrt3\left(\frac{\sqrt3}{2}\right)
\displaystyle =4-3=1.
\displaystyle \therefore a=1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In a }\triangle ABC,\text{ if }\cos A=\frac{\sin B}{2\sin C},\text{ then show that }c=a.
\displaystyle \text{Answer:}
\displaystyle \text{By the sine rule, }\frac{\sin B}{\sin C}=\frac{b}{c}.
\displaystyle \therefore \cos A=\frac{b}{2c}.
\displaystyle \text{By the cosine rule, }\cos A=\frac{b^2+c^2-a^2}{2bc}.
\displaystyle \therefore \frac{b^2+c^2-a^2}{2bc}=\frac{b}{2c}.
\displaystyle \Rightarrow b^2+c^2-a^2=b^2
\displaystyle \Rightarrow c^2=a^2.
\displaystyle \text{Since }a\text{ and }c\text{ are sides of a triangle, }a,c>0.
\displaystyle \therefore c=a.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In a }\triangle ABC,\text{ if }b=20,\ c=21\text{ and }\sin A=\frac35,\text{ find }a.
\displaystyle \text{Answer:}
\displaystyle \sin A=\frac35\Rightarrow \cos A=\pm\sqrt{1-\frac9{25}}=\pm\frac45.
\displaystyle \text{By the cosine rule, }a^2=b^2+c^2-2bc\cos A.
\displaystyle \text{If }A\text{ is acute, }\cos A=\frac45.
\displaystyle a^2=20^2+21^2-2(20)(21)\left(\frac45\right)
\displaystyle =400+441-672=169.
\displaystyle \therefore a=13.
\displaystyle \text{If }A\text{ is obtuse, }\cos A=-\frac45.
\displaystyle a^2=20^2+21^2+2(20)(21)\left(\frac45\right)
\displaystyle =400+441+672=1513.
\displaystyle \therefore a=\sqrt{1513}.
\displaystyle \therefore a=13\text{ or }\sqrt{1513}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In a }\triangle ABC,\text{ if }\sin A\text{ and }\sin B\text{ are the roots of the equation}
\displaystyle c^2x^2-c(a+b)x+ab=0,\text{ then find }\angle C.
\displaystyle \text{Answer:}
\displaystyle \text{Sum of the roots}=\frac{c(a+b)}{c^2}=\frac{a+b}{c}.
\displaystyle \therefore \sin A+\sin B=\frac{a+b}{c}.
\displaystyle \text{By the sine rule, }\frac{a}{c}=\frac{\sin A}{\sin C},\quad\frac{b}{c}=\frac{\sin B}{\sin C}.
\displaystyle \therefore \frac{a+b}{c}=\frac{\sin A+\sin B}{\sin C}.
\displaystyle \therefore \sin A+\sin B=\frac{\sin A+\sin B}{\sin C}.
\displaystyle \Rightarrow \sin C=1.
\displaystyle \therefore C=90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In }\triangle ABC,\text{ if }a=8,\ b=10,\ c=12\text{ and }C=\lambda A,
\displaystyle \text{find the value of }\lambda.
\displaystyle \text{Answer:}
\displaystyle \text{By the cosine rule, }\cos A=\frac{b^2+c^2-a^2}{2bc}.
\displaystyle =\frac{10^2+12^2-8^2}{2(10)(12)}=\frac{180}{240}=\frac34.
\displaystyle \therefore \cos2A=2\cos^2A-1
\displaystyle =2\left(\frac34\right)^2-1=\frac18.
\displaystyle \text{Also, }\cos C=\frac{a^2+b^2-c^2}{2ab}.
\displaystyle =\frac{8^2+10^2-12^2}{2(8)(10)}=\frac{20}{160}=\frac18.
\displaystyle \therefore \cos C=\cos2A.
\displaystyle \text{Since }0<A,C<\pi,\text{ we get }C=2A.
\displaystyle \text{But }C=\lambda A.
\displaystyle \therefore \lambda A=2A\Rightarrow\lambda=2.
\displaystyle \therefore \lambda=2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If the sides of a triangle are proportional to }2,\sqrt6\text{ and }\sqrt3-1,
\displaystyle \text{find the measure of its greatest angle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the sides be }a=2,\quad b=\sqrt6,\quad c=\sqrt3-1.
\displaystyle \text{Since }\sqrt6\text{ is the greatest side, }B\text{ is the greatest angle.}
\displaystyle \text{By the cosine rule, }\cos B=\frac{a^2+c^2-b^2}{2ac}.
\displaystyle =\frac{4+(\sqrt3-1)^2-6}{4(\sqrt3-1)}
\displaystyle =\frac{4+(4-2\sqrt3)-6}{4(\sqrt3-1)}
\displaystyle =\frac{2-2\sqrt3}{4(\sqrt3-1)}=-\frac12.
\displaystyle \therefore B=120^\circ.
\displaystyle \therefore \text{The greatest angle is }120^\circ.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If in a }\triangle ABC,\ \frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c},
\displaystyle \text{then find the measures of angles }A,\ B,\ C.
\displaystyle \text{Answer:}
\displaystyle \text{By the sine rule, }\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k.
\displaystyle \therefore a=k\sin A,\quad b=k\sin B,\quad c=k\sin C.
\displaystyle \text{Given, }\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}.
\displaystyle \therefore \frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}=\frac{\cos C}{\sin C}.
\displaystyle \therefore \cot A=\cot B=\cot C.
\displaystyle \therefore A=B=C.
\displaystyle \text{Also, }A+B+C=180^\circ.
\displaystyle \therefore 3A=180^\circ\Rightarrow A=60^\circ.
\displaystyle \therefore A=B=C=60^\circ.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In any triangle }ABC,\text{ find the value of}
\displaystyle a\sin(B-C)+b\sin(C-A)+c\sin(A-B).
\displaystyle \text{Answer:}
\displaystyle \text{By the sine rule, let }a=k\sin A,\quad b=k\sin B,\quad c=k\sin C.
\displaystyle \text{Required expression}
\displaystyle =k\{\sin A\sin(B-C)+\sin B\sin(C-A)+\sin C\sin(A-B)\}.
\displaystyle =k\{\sin A(\sin B\cos C-\cos B\sin C)
\displaystyle \quad+\sin B(\sin C\cos A-\cos C\sin A)
\displaystyle \quad+\sin C(\sin A\cos B-\cos A\sin B)\}.
\displaystyle \text{All the terms cancel in pairs.}
\displaystyle \therefore \text{Required value}=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In any }\triangle ABC,\text{ find the value of }\Sigma a(\sin B-\sin C).
\displaystyle \text{Answer:}
\displaystyle \Sigma a(\sin B-\sin C)
\displaystyle =a(\sin B-\sin C)+b(\sin C-\sin A)+c(\sin A-\sin B).
\displaystyle \text{By the sine rule, let }a=k\sin A,\quad b=k\sin B,\quad c=k\sin C.
\displaystyle \therefore \Sigma a(\sin B-\sin C)
\displaystyle =k\{\sin A(\sin B-\sin C)+\sin B(\sin C-\sin A)
\displaystyle \quad+\sin C(\sin A-\sin B)\}.
\displaystyle =k\{\sin A\sin B-\sin A\sin C+\sin B\sin C
\displaystyle \quad-\sin A\sin B+\sin A\sin C-\sin B\sin C\}
\displaystyle =0.
\displaystyle \therefore \text{Required value}=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write the number of solutions of the equation}
\displaystyle \tan x+\sec x=2\cos x\text{ in the interval }[0,2\pi].
\displaystyle \text{Answer:}
\displaystyle \tan x+\sec x=2\cos x
\displaystyle \frac{\sin x+1}{\cos x}=2\cos x,\qquad \cos x\ne0.
\displaystyle \therefore \sin x+1=2\cos^2x=2(1-\sin^2x).
\displaystyle \therefore 2\sin^2x+\sin x-1=0.
\displaystyle \therefore (2\sin x-1)(\sin x+1)=0.
\displaystyle \therefore \sin x=\frac12\quad\text{or}\quad\sin x=-1.
\displaystyle \sin x=\frac12\Rightarrow x=\frac{\pi}{6},\frac{5\pi}{6}.
\displaystyle \sin x=-1\Rightarrow x=\frac{3\pi}{2},\text{ but }\cos\frac{3\pi}{2}=0.
\displaystyle \therefore \text{the number of solutions is }2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Write the number of solutions of the equation}
\displaystyle 4\sin x-3\cos x=7.
\displaystyle \text{Answer:}
\displaystyle \text{For }a\sin x+b\cos x,\text{ the maximum value is }\sqrt{a^2+b^2}.
\displaystyle \therefore |4\sin x-3\cos x|\le\sqrt{4^2+(-3)^2}=5.
\displaystyle \text{Since }7>5,\text{ the equation has no solution.}
\displaystyle \therefore \text{the number of solutions is }0.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Write the general solution of }\tan^2 2x=1.
\displaystyle \text{Answer:}
\displaystyle \tan^2 2x=1
\displaystyle \therefore \tan 2x=\pm1.
\displaystyle \therefore 2x=n\pi+\frac{\pi}{4},\qquad n\in Z.
\displaystyle \therefore x=\frac{n\pi}{2}+\frac{\pi}{8},\qquad n\in Z.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Write the set of values of }a\text{ for which the equation}
\displaystyle \sqrt3\sin x-\cos x=a\text{ has no solution.}
\displaystyle \text{Answer:}
\displaystyle |\sqrt3\sin x-\cos x|\le\sqrt{(\sqrt3)^2+(-1)^2}=2.
\displaystyle \therefore -2\le\sqrt3\sin x-\cos x\le2.
\displaystyle \therefore \sqrt3\sin x-\cos x=a\text{ has no solution if }|a|>2.
\displaystyle \therefore a\in(-\infty,-2)\cup(2,\infty).
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\cos x=k\text{ has exactly one solution in }[0,2\pi],
\displaystyle \text{then write the value(s) of }k.
\displaystyle \text{Answer:}
\displaystyle \text{For }-1<k<1,\ \cos x=k\text{ has two solutions in }[0,2\pi].
\displaystyle \text{For }k=1,\ x=0,2\pi,\text{ so there are two solutions.}
\displaystyle \text{For }k=-1,\ x=\pi,\text{ so there is exactly one solution.}
\displaystyle \therefore k=-1.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Write the number of points of intersection of the curves}
\displaystyle 2y=1\text{ and }y=\cos x,\ 0\le x\le2\pi.
\displaystyle \text{Answer:}
\displaystyle 2y=1\Rightarrow y=\frac12.
\displaystyle \therefore \cos x=\frac12.
\displaystyle \text{For }0\le x\le2\pi,\ x=\frac{\pi}{3},\frac{5\pi}{3}.
\displaystyle \therefore \text{the number of points of intersection is }2.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Write the values of }x\text{ in }[0,\pi]\text{ for which }\sin 2x,
\displaystyle \frac12\text{ and }\cos 2x\text{ are in A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\sin 2x,\frac12,\cos 2x\text{ are in A.P.,}
\displaystyle 2\left(\frac12\right)=\sin 2x+\cos 2x.
\displaystyle \therefore \sin 2x+\cos 2x=1.
\displaystyle \therefore \sin 2x=1-\cos 2x.
\displaystyle \therefore 2\sin x\cos x=2\sin^2x.
\displaystyle \therefore 2\sin x(\cos x-\sin x)=0.
\displaystyle \therefore \sin x=0\quad\text{or}\quad\sin x=\cos x.
\displaystyle \text{For }x\in[0,\pi],\quad x=0,\frac{\pi}{4},\pi.
\displaystyle \therefore x=0,\frac{\pi}{4},\pi.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Write the number of points of intersection of the curves}
\displaystyle 2y=-1\text{ and }y=\mathrm{cosec}\,x.
\displaystyle \text{Answer:}
\displaystyle 2y=-1\Rightarrow y=-\frac12.
\displaystyle \text{But for real }x,\quad |\mathrm{cosec}\,x|\ge1.
\displaystyle \therefore \mathrm{cosec}\,x=-\frac12\text{ has no real solution.}
\displaystyle \therefore \text{the number of points of intersection is }0.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Write the solution set of the equation}
\displaystyle (2\cos x+1)(4\cos x+5)=0\text{ in the interval }[0,2\pi].
\displaystyle \text{Answer:}
\displaystyle (2\cos x+1)(4\cos x+5)=0
\displaystyle \therefore \cos x=-\frac12\quad\text{or}\quad\cos x=-\frac54.
\displaystyle \cos x=-\frac54\text{ is not possible since }-1\le\cos x\le1.
\displaystyle \therefore \cos x=-\frac12.
\displaystyle \text{For }x\in[0,2\pi],\quad x=\frac{2\pi}{3},\frac{4\pi}{3}.
\displaystyle \therefore \text{the solution set is }\left\{\frac{2\pi}{3},\frac{4\pi}{3}\right\}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Write the number of values of }x\text{ in }[0,2\pi]
\displaystyle \text{that satisfy the equation }\sin^2x-\cos x=\frac14.
\displaystyle \text{Answer:}
\displaystyle \sin^2x-\cos x=\frac14
\displaystyle \therefore 1-\cos^2x-\cos x=\frac14.
\displaystyle \therefore 4\cos^2x+4\cos x-3=0.
\displaystyle \therefore (2\cos x-1)(2\cos x+3)=0.
\displaystyle \therefore \cos x=\frac12\quad\text{or}\quad\cos x=-\frac32.
\displaystyle \cos x=-\frac32\text{ is not possible since }-1\le\cos x\le1.
\displaystyle \therefore \cos x=\frac12.
\displaystyle \text{For }x\in[0,2\pi],\quad x=\frac{\pi}{3},\frac{5\pi}{3}.
\displaystyle \therefore \text{the number of values of }x\text{ is }2.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }3\tan\left(x-\frac{\pi}{12}\right)=
\displaystyle \tan\left(x+\frac{\pi}{12}\right),\ 0<x<\frac{\pi}{2},\text{ find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=x-\frac{\pi}{12}.
\displaystyle \therefore x+\frac{\pi}{12}=A+\frac{\pi}{6}.
\displaystyle \therefore 3\tan A=\tan\left(A+\frac{\pi}{6}\right).
\displaystyle \text{Let }\tan A=t.
\displaystyle \therefore 3t=\frac{t+\frac{1}{\sqrt3}}{1-\frac{t}{\sqrt3}}.
\displaystyle \therefore 3t(\sqrt3-t)=\sqrt3t+1.
\displaystyle \therefore 3t^2-2\sqrt3t+1=0.
\displaystyle \therefore (\sqrt3t-1)^2=0.
\displaystyle \therefore t=\frac{1}{\sqrt3}.
\displaystyle \therefore \tan A=\tan\frac{\pi}{6}.
\displaystyle \therefore A=n\pi+\frac{\pi}{6},\qquad n\in Z.
\displaystyle \therefore x=n\pi+\frac{\pi}{6}+\frac{\pi}{12}
\displaystyle =n\pi+\frac{\pi}{4}.
\displaystyle \text{Since }0<x<\frac{\pi}{2},\quad x=\frac{\pi}{4}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }2\sin^2x=3\cos x,\text{ where }0\le x\le2\pi,
\displaystyle \text{then find the value of }x.
\displaystyle \text{Answer:}
\displaystyle 2\sin^2x=3\cos x
\displaystyle \therefore 2(1-\cos^2x)=3\cos x.
\displaystyle \therefore 2\cos^2x+3\cos x-2=0.
\displaystyle \therefore (2\cos x-1)(\cos x+2)=0.
\displaystyle \therefore \cos x=\frac12\quad\text{or}\quad\cos x=-2.
\displaystyle \cos x=-2\text{ is not possible since }-1\le\cos x\le1.
\displaystyle \therefore \cos x=\frac12.
\displaystyle \text{For }0\le x\le2\pi,\quad x=\frac{\pi}{3},\frac{5\pi}{3}.
\displaystyle \therefore x=\frac{\pi}{3},\frac{5\pi}{3}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }\sec x\cos5x+1=0,\text{ where }0<x\le\frac{\pi}{2},
\displaystyle \text{find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \sec x\cos5x+1=0,\qquad \cos x\ne0.
\displaystyle \therefore \frac{\cos5x}{\cos x}+1=0.
\displaystyle \therefore \cos5x+\cos x=0.
\displaystyle \therefore 2\cos3x\cos2x=0.
\displaystyle \therefore \cos3x=0\quad\text{or}\quad\cos2x=0.
\displaystyle \cos3x=0\Rightarrow 3x=(2n+1)\frac{\pi}{2}.
\displaystyle \therefore x=(2n+1)\frac{\pi}{6}.
\displaystyle \text{In }0<x\le\frac{\pi}{2},\quad x=\frac{\pi}{6},\frac{\pi}{2}.
\displaystyle \cos2x=0\Rightarrow 2x=(2n+1)\frac{\pi}{2}.
\displaystyle \therefore x=(2n+1)\frac{\pi}{4}.
\displaystyle \text{In }0<x\le\frac{\pi}{2},\quad x=\frac{\pi}{4}.
\displaystyle \text{But }x=\frac{\pi}{2}\text{ is rejected since }\sec\frac{\pi}{2}\text{ is undefined.}
\displaystyle \therefore x=\frac{\pi}{6},\frac{\pi}{4}.
\displaystyle \\

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