\displaystyle \textbf{Question 1: }\text{Find the general solutions of the following equations:}
\displaystyle \text{(i) }\sin x=\frac{1}{2}\qquad \text{(ii) }\cos x=-\frac{\sqrt{3}}{2}\qquad
\displaystyle \text{(iii) }\mathrm{cosec}\,x=-\sqrt{2}
\displaystyle \text{(iv) }\sec x=\sqrt{2}\qquad \text{(v) }\tan x=-\frac{1}{\sqrt{3}}\qquad
\displaystyle \text{(vi) }\sqrt{3}\sec x=2.
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\sin x=\frac{1}{2}
\displaystyle \Rightarrow \sin x=\sin\frac{\pi}{6}
\displaystyle \therefore x=n\pi+(-1)^n\frac{\pi}{6},\quad n\in\mathbb Z.

\displaystyle \text{(ii) }\cos x=-\frac{\sqrt{3}}{2}
\displaystyle \Rightarrow \cos x=\cos\frac{5\pi}{6}
\displaystyle \therefore x=2n\pi\pm\frac{5\pi}{6},\quad n\in\mathbb Z.

\displaystyle \text{(iii) }\mathrm{cosec}\,x=-\sqrt{2}
\displaystyle \Rightarrow \frac{1}{\sin x}=-\sqrt{2}
\displaystyle \Rightarrow \sin x=-\frac{1}{\sqrt{2}}=-\sin\frac{\pi}{4}
\displaystyle \Rightarrow \sin x=\sin\left(-\frac{\pi}{4}\right)
\displaystyle \therefore x=n\pi+(-1)^n\left(-\frac{\pi}{4}\right),\quad n\in\mathbb Z
\displaystyle \Rightarrow x=n\pi+(-1)^{n+1}\frac{\pi}{4},\quad n\in\mathbb Z.

\displaystyle \text{(iv) }\sec x=\sqrt{2}
\displaystyle \Rightarrow \frac{1}{\cos x}=\sqrt{2}
\displaystyle \Rightarrow \cos x=\frac{1}{\sqrt{2}}=\cos\frac{\pi}{4}
\displaystyle \therefore x=2n\pi\pm\frac{\pi}{4},\quad n\in\mathbb Z.

\displaystyle \text{(v) }\tan x=-\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \tan x=-\tan\frac{\pi}{6}
\displaystyle \Rightarrow \tan x=\tan\left(-\frac{\pi}{6}\right)
\displaystyle \therefore x=n\pi-\frac{\pi}{6},\quad n\in\mathbb Z.

\displaystyle \text{(vi) }\sqrt{3}\sec x=2
\displaystyle \Rightarrow \frac{\sqrt{3}}{\cos x}=2
\displaystyle \Rightarrow \cos x=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}
\displaystyle \therefore x=2n\pi\pm\frac{\pi}{6},\quad n\in\mathbb Z.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the general solutions of the following equations:}
\displaystyle \text{(i) }\sin2x=\frac{\sqrt{3}}{2}\qquad\text{(ii) }\cos3x=\frac{1}{2}\qquad\text{(iii) }\sin9x=\sin x
\displaystyle \text{(iv) }\sin2x=\cos3x\qquad\text{(v) }\tan x+\cot2x=0\qquad\text{(vi) }\tan3x=\cot x
\displaystyle \text{(vii) }\tan2x\tan x=1\qquad\text{(viii) }\tan mx+\cot nx=0
\displaystyle \text{(ix) }\tan px=\cot qx\qquad\text{(x) }\sin2x+\cos x=0
\displaystyle \text{(xi) }\sin x=\tan x\qquad\text{(xii) }\sin3x+\cos2x=0.
\displaystyle \text{Answer:}

\displaystyle \text{i) }\sin2x=\frac{\sqrt{3}}{2}
\displaystyle \Rightarrow \sin2x=\sin\frac{\pi}{3}
\displaystyle \Rightarrow 2x=n\pi+(-1)^n\frac{\pi}{3};\ n\in Z
\displaystyle \Rightarrow x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{6};\ n\in Z

\displaystyle \text{ii) }\cos3x=\frac{1}{2}
\displaystyle \Rightarrow \cos3x=\cos\frac{\pi}{3}
\displaystyle \Rightarrow 3x=2n\pi\pm\frac{\pi}{3};\ n\in Z
\displaystyle \Rightarrow x=\frac{2n\pi}{3}\pm\frac{\pi}{9};\ n\in Z

\displaystyle \text{iii) }\sin9x=\sin x
\displaystyle \Rightarrow \sin9x-\sin x=0
\displaystyle \Rightarrow 2\cos5x\sin4x=0
\displaystyle \therefore \cos5x=0\quad\text{or}\quad\sin4x=0.
\displaystyle \cos5x=0
\displaystyle \Rightarrow \cos5x=\cos\frac{\pi}{2}
\displaystyle \Rightarrow 5x=\frac{(2n+1)\pi}{2};\ n\in Z
\displaystyle \Rightarrow x=\frac{(2n+1)\pi}{10};\ n\in Z
\displaystyle \sin4x=0
\displaystyle \Rightarrow \sin4x=\sin0
\displaystyle \Rightarrow 4x=n\pi+(-1)^n(0);\ n\in Z
\displaystyle \Rightarrow x=\frac{n\pi}{4};\ n\in Z

\displaystyle \text{iv) }\sin2x=\cos3x
\displaystyle \Rightarrow \cos\left(\frac{\pi}{2}-2x\right)=\cos3x
\displaystyle \Rightarrow 3x=2n\pi\pm\left(\frac{\pi}{2}-2x\right),\quad n\in Z
\displaystyle \text{Considering the positive sign,}
\displaystyle 3x=2n\pi+\left(\frac{\pi}{2}-2x\right)
\displaystyle \Rightarrow 5x=2n\pi+\frac{\pi}{2}
\displaystyle \Rightarrow 5x=\frac{(4n+1)\pi}{2}
\displaystyle \Rightarrow x=\frac{(4n+1)\pi}{10},\quad n\in Z
\displaystyle \text{Considering the negative sign,}
\displaystyle 3x=2n\pi-\left(\frac{\pi}{2}-2x\right)
\displaystyle \Rightarrow 3x=2n\pi-\frac{\pi}{2}+2x
\displaystyle \Rightarrow x=2n\pi-\frac{\pi}{2}
\displaystyle \Rightarrow x=\frac{(4n-1)\pi}{2},\quad n\in Z
\displaystyle \therefore x=\frac{(4n+1)\pi}{10}\quad\text{or}\quad x=\frac{(4n-1)\pi}{2},\quad n\in Z

\displaystyle \text{v) }\tan x+\cot2x=0
\displaystyle \Rightarrow \frac{\sin x}{\cos x}+\frac{\cos2x}{\sin2x}=0
\displaystyle \text{Here, }\cos x\ne0\text{ and }\sin2x\ne0.
\displaystyle \Rightarrow \frac{\sin x\sin2x+\cos x\cos2x}{\cos x\sin2x}=0
\displaystyle \Rightarrow \sin x\sin2x+\cos x\cos2x=0
\displaystyle \Rightarrow \cos(2x-x)=0
\displaystyle \Rightarrow \cos x=0
\displaystyle \Rightarrow x=\frac{(2n+1)\pi}{2},\quad n\in Z
\displaystyle \text{But for these values, }\tan x\text{ and }\cot2x\text{ are not defined.}
\displaystyle \therefore \text{There is no real solution.}

\displaystyle \text{vi) }\tan3x=\cot x
\displaystyle \Rightarrow \tan3x=\tan\left(\frac{\pi}{2}-x\right)
\displaystyle \Rightarrow 3x=n\pi+\left(\frac{\pi}{2}-x\right),\quad n\in Z
\displaystyle \Rightarrow 4x=n\pi+\frac{\pi}{2}
\displaystyle \Rightarrow 4x=\frac{(2n+1)\pi}{2}
\displaystyle \Rightarrow x=\frac{(2n+1)\pi}{8},\quad n\in Z

\displaystyle \text{vii) }\tan2x\tan x=1
\displaystyle \Rightarrow \tan2x=\frac{1}{\tan x}=\cot x
\displaystyle \Rightarrow \tan2x=\tan\left(\frac{\pi}{2}-x\right)
\displaystyle \Rightarrow 2x=n\pi+\left(\frac{\pi}{2}-x\right),\quad n\in Z
\displaystyle \Rightarrow 3x=\frac{(2n+1)\pi}{2}
\displaystyle \Rightarrow x=\frac{(2n+1)\pi}{6}
\displaystyle \text{For }n=3k+1,\quad x=\frac{(2k+1)\pi}{2},
\displaystyle \text{for which }\tan x\text{ is not defined.}
\displaystyle \therefore x=k\pi+\frac{\pi}{6}\quad\text{or}\quad x=k\pi+\frac{5\pi}{6},\quad k\in Z

\displaystyle \text{viii) }\tan mx+\cot nx=0
\displaystyle \Rightarrow \tan mx=-\cot nx
\displaystyle \Rightarrow \tan mx=\tan\left(nx+\frac{\pi}{2}\right)
\displaystyle \Rightarrow mx=k\pi+nx+\frac{\pi}{2},\quad k\in Z
\displaystyle \Rightarrow (m-n)x=\frac{(2k+1)\pi}{2}
\displaystyle \therefore x=\frac{(2k+1)\pi}{2(m-n)},\quad k\in Z,\quad m\ne n,
\displaystyle \text{provided }\sin nx\ne0.
\displaystyle \text{If }m=n,\text{ there is no solution.}

\displaystyle \text{ix) }\tan px=\cot qx
\displaystyle \Rightarrow \tan px=\tan\left(\frac{\pi}{2}-qx\right)
\displaystyle \Rightarrow px=n\pi+\frac{\pi}{2}-qx,\quad n\in Z
\displaystyle \Rightarrow (p+q)x=\frac{(2n+1)\pi}{2}
\displaystyle \therefore x=\frac{(2n+1)\pi}{2(p+q)},\quad n\in Z,\quad p+q\ne0,
\displaystyle \text{provided }\sin qx\ne0.
\displaystyle \text{If }p+q=0,\text{ there is no solution.}

\displaystyle \text{x) }\sin2x+\cos x=0
\displaystyle \Rightarrow 2\sin x\cos x+\cos x=0
\displaystyle \Rightarrow \cos x(2\sin x+1)=0
\displaystyle \therefore \cos x=0\quad\text{or}\quad\sin x=-\frac{1}{2}.
\displaystyle \cos x=0
\displaystyle \Rightarrow x=\frac{(2n+1)\pi}{2},\quad n\in Z
\displaystyle \sin x=-\frac{1}{2}
\displaystyle \Rightarrow \sin x=\sin\left(-\frac{\pi}{6}\right)
\displaystyle \Rightarrow x=n\pi+(-1)^n\left(-\frac{\pi}{6}\right),\quad n\in Z
\displaystyle \Rightarrow x=n\pi+(-1)^{n+1}\frac{\pi}{6},\quad n\in Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{2}\quad\text{or}\quad x=n\pi+(-1)^{n+1}\frac{\pi}{6},\quad n\in Z

\displaystyle \text{xi) }\sin x=\tan x
\displaystyle \Rightarrow \sin x=\frac{\sin x}{\cos x},\quad\cos x\ne0
\displaystyle \Rightarrow \sin x\cos x=\sin x
\displaystyle \Rightarrow \sin x(\cos x-1)=0
\displaystyle \therefore \sin x=0\quad\text{or}\quad\cos x=1.
\displaystyle \sin x=0
\displaystyle \Rightarrow x=n\pi,\quad n\in Z
\displaystyle \cos x=1
\displaystyle \Rightarrow x=2m\pi,\quad m\in Z
\displaystyle \text{Since }x=2m\pi\text{ is already included in }x=n\pi,
\displaystyle \therefore x=n\pi,\quad n\in Z

\displaystyle \text{xii) }\sin3x+\cos2x=0
\displaystyle \Rightarrow \cos2x=-\sin3x
\displaystyle \Rightarrow \cos2x=-\cos\left(\frac{\pi}{2}-3x\right)
\displaystyle \Rightarrow \cos2x=\cos\left(\frac{\pi}{2}+3x\right)
\displaystyle \Rightarrow 2x=2n\pi\pm\left(\frac{\pi}{2}+3x\right),\quad n\in Z
\displaystyle \text{Considering the positive sign,}
\displaystyle 2x=2n\pi+\left(\frac{\pi}{2}+3x\right)
\displaystyle \Rightarrow -x=2n\pi+\frac{\pi}{2}
\displaystyle \Rightarrow x=-\frac{(4n+1)\pi}{2},\quad n\in Z
\displaystyle \text{Considering the negative sign,}
\displaystyle 2x=2n\pi-\left(\frac{\pi}{2}+3x\right)
\displaystyle \Rightarrow 5x=2n\pi-\frac{\pi}{2}
\displaystyle \Rightarrow 5x=\frac{(4n-1)\pi}{2}
\displaystyle \Rightarrow x=\frac{(4n-1)\pi}{10},\quad n\in Z
\displaystyle \therefore x=-\frac{(4n+1)\pi}{2}\quad\text{or}\quad x=\frac{(4n-1)\pi}{10},\quad n\in Z

\displaystyle \textbf{Question 3: }\text{Solve the following equations:}
\displaystyle \text{(i) }\sin^2x-\cos x=\frac{1}{4}\qquad\text{(ii) }2\cos^2x-5\cos x+2=0
\displaystyle \text{(iii) }2\sin^2x+\sqrt{3}\cos x+1=0\qquad\text{(iv) }4\sin^2x-8\cos x+1=0
\displaystyle \text{(v) }\tan^2x+(1-\sqrt{3})\tan x-\sqrt{3}=0\qquad\text{(vi) }3\cos^2x-2\sqrt{3}\sin x\cos x-3\sin^2x=0
\displaystyle \text{(vii) }\cos4x=\cos2x

\displaystyle \text{Answer:}

\displaystyle \text{i) }\sin^2x-\cos x=\frac{1}{4}
\displaystyle \Rightarrow 1-\cos^2x-\cos x=\frac{1}{4}
\displaystyle \Rightarrow 4-4\cos^2x-4\cos x=1
\displaystyle \Rightarrow 4\cos^2x+4\cos x-3=0
\displaystyle \Rightarrow (2\cos x-1)(2\cos x+3)=0
\displaystyle \therefore 2\cos x-1=0\quad\text{or}\quad 2\cos x+3=0.
\displaystyle \cos x=\frac{1}{2}\quad\text{or}\quad\cos x=-\frac{3}{2}.
\displaystyle \text{But }\cos x=-\frac{3}{2}\text{ is not possible for real }x.
\displaystyle \therefore \cos x=\frac{1}{2}
\displaystyle \Rightarrow x=2n\pi\pm\frac{\pi}{3},\quad n\in Z

\displaystyle \text{ii) }2\cos^2x-5\cos x+2=0
\displaystyle \Rightarrow (2\cos x-1)(\cos x-2)=0
\displaystyle \therefore 2\cos x-1=0\quad\text{or}\quad\cos x-2=0.
\displaystyle \cos x=\frac{1}{2}\quad\text{or}\quad\cos x=2.
\displaystyle \text{But }\cos x=2\text{ is not possible for real }x.
\displaystyle \therefore \cos x=\frac{1}{2}
\displaystyle \Rightarrow x=2n\pi\pm\frac{\pi}{3},\quad n\in Z

\displaystyle \text{iii) }2\sin^2x+\sqrt{3}\cos x+1=0
\displaystyle \Rightarrow 2(1-\cos^2x)+\sqrt{3}\cos x+1=0
\displaystyle \Rightarrow 2-2\cos^2x+\sqrt{3}\cos x+1=0
\displaystyle \Rightarrow 2\cos^2x-\sqrt{3}\cos x-3=0
\displaystyle \Rightarrow (2\cos x+\sqrt{3})(\cos x-\sqrt{3})=0
\displaystyle \therefore 2\cos x+\sqrt{3}=0\quad\text{or}\quad\cos x-\sqrt{3}=0.
\displaystyle \Rightarrow \cos x=-\frac{\sqrt{3}}{2}\quad\text{or}\quad\cos x=\sqrt{3}.
\displaystyle \text{But }\cos x=\sqrt{3}\text{ is not possible for real }x.
\displaystyle \therefore \cos x=-\frac{\sqrt{3}}{2}
\displaystyle \Rightarrow \cos x=\cos\frac{5\pi}{6}
\displaystyle \therefore x=2n\pi\pm\frac{5\pi}{6},\quad n\in Z

\displaystyle \text{iv) }4\sin^2x-8\cos x+1=0
\displaystyle \Rightarrow 4(1-\cos^2x)-8\cos x+1=0
\displaystyle \Rightarrow 4-4\cos^2x-8\cos x+1=0
\displaystyle \Rightarrow 4\cos^2x+8\cos x-5=0
\displaystyle \Rightarrow (2\cos x-1)(2\cos x+5)=0
\displaystyle \therefore 2\cos x-1=0\quad\text{or}\quad2\cos x+5=0.
\displaystyle \Rightarrow \cos x=\frac{1}{2}\quad\text{or}\quad\cos x=-\frac{5}{2}.
\displaystyle \text{But }\cos x=-\frac{5}{2}\text{ is not possible for real }x.
\displaystyle \therefore \cos x=\frac{1}{2}
\displaystyle \Rightarrow \cos x=\cos\frac{\pi}{3}
\displaystyle \therefore x=2n\pi\pm\frac{\pi}{3},\quad n\in Z

\displaystyle \text{v) }\tan^2x+(1-\sqrt{3})\tan x-\sqrt{3}=0
\displaystyle \Rightarrow (\tan x+1)(\tan x-\sqrt{3})=0
\displaystyle \therefore \tan x=-1\quad\text{or}\quad\tan x=\sqrt{3}.
\displaystyle \tan x=-1
\displaystyle \Rightarrow \tan x=\tan\left(-\frac{\pi}{4}\right)
\displaystyle \Rightarrow x=n\pi-\frac{\pi}{4},\quad n\in Z
\displaystyle \tan x=\sqrt{3}
\displaystyle \Rightarrow \tan x=\tan\frac{\pi}{3}
\displaystyle \Rightarrow x=n\pi+\frac{\pi}{3},\quad n\in Z
\displaystyle \therefore x=n\pi-\frac{\pi}{4}\quad\text{or}\quad x=n\pi+\frac{\pi}{3},\quad n\in Z

\displaystyle \text{vi) }3\cos^2x-2\sqrt{3}\sin x\cos x-3\sin^2x=0
\displaystyle \text{For }\cos x=0,\text{ the left-hand side is }-3\ne0.
\displaystyle \therefore \cos x\ne0.
\displaystyle \text{Dividing throughout by }\cos^2x,
\displaystyle 3-2\sqrt{3}\tan x-3\tan^2x=0
\displaystyle \Rightarrow 3\tan^2x+2\sqrt{3}\tan x-3=0
\displaystyle \Rightarrow (\sqrt{3}\tan x-1)(\sqrt{3}\tan x+3)=0
\displaystyle \therefore \tan x=\frac{1}{\sqrt{3}}\quad\text{or}\quad\tan x=-\sqrt{3}.
\displaystyle \tan x=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \tan x=\tan\frac{\pi}{6}
\displaystyle \Rightarrow x=n\pi+\frac{\pi}{6},\quad n\in Z
\displaystyle \tan x=-\sqrt{3}
\displaystyle \Rightarrow \tan x=\tan\left(-\frac{\pi}{3}\right)
\displaystyle \Rightarrow x=n\pi-\frac{\pi}{3},\quad n\in Z
\displaystyle \therefore x=n\pi+\frac{\pi}{6}\quad\text{or}\quad x=n\pi-\frac{\pi}{3},\quad n\in Z

\displaystyle \text{vii) }\cos4x=\cos2x
\displaystyle \Rightarrow 4x=2n\pi\pm2x,\quad n\in Z
\displaystyle \text{Considering the positive sign,}
\displaystyle 4x=2n\pi+2x
\displaystyle \Rightarrow 2x=2n\pi
\displaystyle \Rightarrow x=n\pi,\quad n\in Z
\displaystyle \text{Considering the negative sign,}
\displaystyle 4x=2n\pi-2x
\displaystyle \Rightarrow 6x=2n\pi
\displaystyle \Rightarrow x=\frac{n\pi}{3},\quad n\in Z
\displaystyle \text{Since }x=n\pi\text{ is included in }x=\frac{n\pi}{3},
\displaystyle \therefore x=\frac{n\pi}{3},\quad n\in Z

\displaystyle \textbf{Question 4: }\text{Solve the following equations:}
\displaystyle \text{i) }\cos x+\cos2x+\cos3x=0\hspace{1cm}\text{ii) }\cos x+\cos3x-\cos2x=0
\displaystyle \text{iii) }\sin x+\sin5x=\sin3x\hspace{1cm}\text{iv) }\cos x\cos2x\cos3x=\frac{1}{4}
\displaystyle \text{v) }\cos x+\sin x=\cos2x+\sin2x\hspace{1cm}\text{vi) }\sin x+\sin2x+\sin3x=0
\displaystyle \text{vii) }\sin x+\sin2x+\sin3x+\sin4x=0
\displaystyle \text{viii) }\sin3x-\sin x=4\cos^2x-2
\displaystyle \text{ix) }\sin2x-\sin4x+\sin6x=0
\displaystyle \text{Answer:}

\displaystyle \text{i) }\cos x+\cos2x+\cos3x=0
\displaystyle \text{Using }\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle \cos x+\cos3x=2\cos2x\cos x
\displaystyle \therefore 2\cos2x\cos x+\cos2x=0
\displaystyle \cos2x(2\cos x+1)=0
\displaystyle \therefore \cos2x=0\quad\text{or}\quad 2\cos x+1=0
\displaystyle \text{Case I: }\cos2x=0
\displaystyle 2x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{4},\quad n\in\mathbb Z
\displaystyle \text{Case II: }2\cos x+1=0
\displaystyle \cos x=-\frac{1}{2}
\displaystyle \therefore x=2n\pi\pm\frac{2\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{4}\quad\text{or}\quad x=2n\pi\pm\frac{2\pi}{3},\quad n\in\mathbb Z

\displaystyle \text{ii) }\cos x+\cos3x-\cos2x=0
\displaystyle \text{Using }\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle \cos x+\cos3x=2\cos2x\cos x
\displaystyle \therefore 2\cos2x\cos x-\cos2x=0
\displaystyle \cos2x(2\cos x-1)=0
\displaystyle \therefore \cos2x=0\quad\text{or}\quad 2\cos x-1=0
\displaystyle \text{Case I: }\cos2x=0
\displaystyle 2x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{4},\quad n\in\mathbb Z
\displaystyle \text{Case II: }2\cos x-1=0
\displaystyle \cos x=\frac{1}{2}
\displaystyle \therefore x=2n\pi\pm\frac{\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{4}\quad\text{or}\quad x=2n\pi\pm\frac{\pi}{3},\quad n\in\mathbb Z

\displaystyle \text{iii) }\sin x+\sin5x=\sin3x
\displaystyle \text{Using }\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle \sin x+\sin5x=2\sin3x\cos2x
\displaystyle \therefore 2\sin3x\cos2x=\sin3x
\displaystyle \sin3x(2\cos2x-1)=0
\displaystyle \therefore \sin3x=0\quad\text{or}\quad 2\cos2x-1=0
\displaystyle \text{Case I: }\sin3x=0
\displaystyle 3x=n\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{n\pi}{3},\quad n\in\mathbb Z
\displaystyle \text{Case II: }2\cos2x-1=0
\displaystyle \cos2x=\frac{1}{2}
\displaystyle 2x=2n\pi\pm\frac{\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=n\pi\pm\frac{\pi}{6},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{n\pi}{3}\quad\text{or}\quad x=n\pi\pm\frac{\pi}{6},\quad n\in\mathbb Z

\displaystyle \text{iv) }\cos x\cos2x\cos3x=\frac{1}{4}
\displaystyle \text{Given, }\cos x\cos2x\cos3x=\frac{1}{4}
\displaystyle \therefore 4\cos x\cos2x\cos3x=1
\displaystyle \text{Using }2\cos A\cos B=\cos(A+B)+\cos(A-B),
\displaystyle 2\cos x\cos3x=\cos4x+\cos2x
\displaystyle \therefore 4\cos x\cos2x\cos3x=2\cos2x(\cos4x+\cos2x)
\displaystyle =2\cos2x\cos4x+2\cos^22x
\displaystyle =(\cos6x+\cos2x)+(1+\cos4x)
\displaystyle \therefore \cos2x+\cos4x+\cos6x+1=1
\displaystyle \cos2x+\cos4x+\cos6x=0
\displaystyle (\cos2x+\cos6x)+\cos4x=0
\displaystyle 2\cos4x\cos2x+\cos4x=0
\displaystyle \cos4x(2\cos2x+1)=0
\displaystyle \therefore \cos4x=0\quad\text{or}\quad 2\cos2x+1=0
\displaystyle \text{Case I: }\cos4x=0
\displaystyle 4x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{8},\quad n\in\mathbb Z
\displaystyle \text{Case II: }2\cos2x+1=0
\displaystyle \cos2x=-\frac{1}{2}
\displaystyle 2x=2n\pi\pm\frac{2\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=n\pi\pm\frac{\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{8}\quad\text{or}\quad x=n\pi\pm\frac{\pi}{3},\quad n\in\mathbb Z

\displaystyle \text{v) }\cos x+\sin x=\cos2x+\sin2x
\displaystyle \cos x-\cos2x+\sin x-\sin2x=0
\displaystyle \text{Using }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \cos x-\cos2x=2\sin\frac{3x}{2}\sin\frac{x}{2}
\displaystyle \text{Also, using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \sin x-\sin2x=-2\cos\frac{3x}{2}\sin\frac{x}{2}
\displaystyle \therefore 2\sin\frac{x}{2}\left(\sin\frac{3x}{2}-\cos\frac{3x}{2}\right)=0
\displaystyle \therefore \sin\frac{x}{2}=0\quad\text{or}\quad \sin\frac{3x}{2}-\cos\frac{3x}{2}=0
\displaystyle \text{Case I: }\sin\frac{x}{2}=0
\displaystyle \frac{x}{2}=n\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=2n\pi,\quad n\in\mathbb Z
\displaystyle \text{Case II: }\sin\frac{3x}{2}=\cos\frac{3x}{2}
\displaystyle \frac{3x}{2}=n\pi+\frac{\pi}{4},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{2n\pi}{3}+\frac{\pi}{6},\quad n\in\mathbb Z
\displaystyle \therefore x=2n\pi\quad\text{or}\quad x=\frac{\pi}{6}+\frac{2n\pi}{3},\quad n\in\mathbb Z

\displaystyle \text{vi) }\sin x+\sin2x+\sin3x=0
\displaystyle \text{Using }\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle \sin x+\sin3x=2\sin2x\cos x
\displaystyle \therefore 2\sin2x\cos x+\sin2x=0
\displaystyle \sin2x(2\cos x+1)=0
\displaystyle \therefore \sin2x=0\quad\text{or}\quad 2\cos x+1=0
\displaystyle \text{Case I: }\sin2x=0
\displaystyle 2x=n\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{n\pi}{2},\quad n\in\mathbb Z
\displaystyle \text{Case II: }2\cos x+1=0
\displaystyle \cos x=-\frac{1}{2}
\displaystyle \therefore x=2n\pi\pm\frac{2\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{n\pi}{2}\quad\text{or}\quad x=2n\pi\pm\frac{2\pi}{3},\quad n\in\mathbb Z

\displaystyle \text{vii) }\sin x+\sin2x+\sin3x+\sin4x=0
\displaystyle (\sin x+\sin4x)+(\sin2x+\sin3x)=0
\displaystyle \text{Using }\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle 2\sin\frac{5x}{2}\cos\frac{3x}{2}+2\sin\frac{5x}{2}\cos\frac{x}{2}=0
\displaystyle 2\sin\frac{5x}{2}\left(\cos\frac{3x}{2}+\cos\frac{x}{2}\right)=0
\displaystyle \text{Using }\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle 4\sin\frac{5x}{2}\cos x\cos\frac{x}{2}=0
\displaystyle \therefore \sin\frac{5x}{2}=0\quad\text{or}\quad\cos x=0\quad\text{or}\quad\cos\frac{x}{2}=0
\displaystyle \text{Case I: }\sin\frac{5x}{2}=0
\displaystyle \frac{5x}{2}=n\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{2n\pi}{5},\quad n\in\mathbb Z
\displaystyle \text{Case II: }\cos x=0
\displaystyle \therefore x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z
\displaystyle \text{Case III: }\cos\frac{x}{2}=0
\displaystyle \frac{x}{2}=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z
\displaystyle \therefore x=(2n+1)\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{2n\pi}{5}\quad\text{or}\quad x=\frac{(2n+1)\pi}{2}\quad\text{or}\quad x=(2n+1)\pi,\quad n\in\mathbb Z

\displaystyle \text{viii) }\sin3x-\sin x=4\cos^2x-2
\displaystyle \text{Using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \sin3x-\sin x=2\cos2x\sin x
\displaystyle \text{Also, }4\cos^2x-2=2(2\cos^2x-1)=2\cos2x
\displaystyle \therefore 2\cos2x\sin x=2\cos2x
\displaystyle 2\cos2x(\sin x-1)=0
\displaystyle \therefore \cos2x=0\quad\text{or}\quad\sin x=1
\displaystyle \text{Case I: }\cos2x=0
\displaystyle 2x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{4},\quad n\in\mathbb Z
\displaystyle \text{Case II: }\sin x=1
\displaystyle \therefore x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{(2n+1)\pi}{4}\quad\text{or}\quad x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z

\displaystyle \text{ix) }\sin2x-\sin4x+\sin6x=0
\displaystyle (\sin2x+\sin6x)-\sin4x=0
\displaystyle \text{Using }\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle 2\sin4x\cos2x-\sin4x=0
\displaystyle \sin4x(2\cos2x-1)=0
\displaystyle \therefore \sin4x=0\quad\text{or}\quad 2\cos2x-1=0
\displaystyle \text{Case I: }\sin4x=0
\displaystyle 4x=n\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{n\pi}{4},\quad n\in\mathbb Z
\displaystyle \text{Case II: }2\cos2x-1=0
\displaystyle \cos2x=\frac{1}{2}
\displaystyle 2x=2n\pi\pm\frac{\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=n\pi\pm\frac{\pi}{6},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{n\pi}{4}\quad\text{or}\quad x=n\pi\pm\frac{\pi}{6},\quad n\in\mathbb Z

\displaystyle \textbf{Question 5: }\text{Solve the following equations:}
\displaystyle \text{i) }\tan x+\tan2x+\tan3x=0
\displaystyle \text{ii) }\tan x+\tan2x=\tan3x
\displaystyle \text{iii) }\tan3x+\tan x=2\tan2x
\displaystyle \text{Answer:}
\displaystyle \text{i) }\tan x+\tan2x+\tan3x=0
\displaystyle \frac{\sin x}{\cos x}+\frac{\sin2x}{\cos2x}+\frac{\sin3x}{\cos3x}=0
\displaystyle \frac{\sin x\cos2x\cos3x+\sin2x\cos x\cos3x+\sin3x\cos x\cos2x}{\cos x\cos2x\cos3x}=0
\displaystyle \sin x\cos2x+\cos x\sin2x=\sin3x
\displaystyle \therefore \sin3x\cos3x+\sin3x\cos x\cos2x=0
\displaystyle \sin3x(\cos3x+\cos x\cos2x)=0
\displaystyle \text{Using }\cos x\cos2x=\frac{\cos3x+\cos x}{2},
\displaystyle \sin3x\left(\cos3x+\frac{\cos3x+\cos x}{2}\right)=0
\displaystyle \sin3x(3\cos3x+\cos x)=0
\displaystyle \therefore \sin3x=0\quad\text{or}\quad 3\cos3x+\cos x=0
\displaystyle \text{Case I: }\sin3x=0
\displaystyle 3x=n\pi
\displaystyle \therefore x=\frac{n\pi}{3}
\displaystyle \text{Case II: }3\cos3x+\cos x=0
\displaystyle 3(4\cos^3x-3\cos x)+\cos x=0
\displaystyle 12\cos^3x-8\cos x=0
\displaystyle 4\cos x(3\cos^2x-2)=0
\displaystyle \cos x=0\quad\text{or}\quad\cos^2x=\frac{2}{3}
\displaystyle \cos x=0\text{ is inadmissible, since }\tan x\text{ is undefined.}
\displaystyle \cos^2x=\frac{2}{3}
\displaystyle \tan^2x=\frac{1-\cos^2x}{\cos^2x}=\frac{1}{2}
\displaystyle \tan x=\pm\frac{1}{\sqrt2}
\displaystyle \therefore x=n\pi\pm\tan^{-1}\frac{1}{\sqrt2}
\displaystyle \therefore x=\frac{n\pi}{3}\quad\text{or}\quad x=n\pi\pm\tan^{-1}\frac{1}{\sqrt2},\quad n\in\mathbb Z

\displaystyle \text{ii) }\tan x+\tan2x=\tan3x
\displaystyle \frac{\sin x}{\cos x}+\frac{\sin2x}{\cos2x}=\frac{\sin3x}{\cos3x}
\displaystyle \frac{\sin x\cos2x+\cos x\sin2x}{\cos x\cos2x}=\frac{\sin3x}{\cos3x}
\displaystyle \frac{\sin3x}{\cos x\cos2x}=\frac{\sin3x}{\cos3x}
\displaystyle \sin3x(\cos3x-\cos x\cos2x)=0
\displaystyle \therefore \sin3x=0\quad\text{or}\quad\cos3x=\cos x\cos2x
\displaystyle \text{Case I: }\sin3x=0
\displaystyle 3x=n\pi
\displaystyle \therefore x=\frac{n\pi}{3}
\displaystyle \text{Case II: }\cos3x=\cos x\cos2x
\displaystyle \cos3x=\frac{\cos3x+\cos x}{2}
\displaystyle \therefore \cos3x=\cos x
\displaystyle 3x=2n\pi\pm x
\displaystyle \therefore x=n\pi\quad\text{or}\quad x=\frac{n\pi}{2}
\displaystyle x=n\pi\text{ is already included in }x=\frac{n\pi}{3}.
\displaystyle \text{For odd }n,\ x=\frac{n\pi}{2}\text{ is inadmissible, since }\tan x\text{ is undefined.}
\displaystyle \text{For even }n,\ x=\frac{n\pi}{2}=k\pi\text{ is already included in }x=\frac{n\pi}{3}.
\displaystyle \therefore x=\frac{n\pi}{3},\quad n\in\mathbb Z

\displaystyle \text{iii) }\tan3x+\tan x=2\tan2x
\displaystyle \frac{\sin3x}{\cos3x}+\frac{\sin x}{\cos x}=2\frac{\sin2x}{\cos2x}
\displaystyle \frac{\sin3x\cos x+\cos3x\sin x}{\cos3x\cos x}=2\frac{\sin2x}{\cos2x}
\displaystyle \frac{\sin4x}{\cos3x\cos x}=2\frac{\sin2x}{\cos2x}
\displaystyle \frac{2\sin2x\cos2x}{\cos3x\cos x}=2\frac{\sin2x}{\cos2x}
\displaystyle 2\sin2x(\cos^22x-\cos3x\cos x)=0
\displaystyle \text{Using }\cos^22x=\frac{1+\cos4x}{2}\text{ and }\cos3x\cos x=\frac{\cos4x+\cos2x}{2},
\displaystyle 2\sin2x\left(\frac{1-\cos2x}{2}\right)=0
\displaystyle \sin2x(1-\cos2x)=0
\displaystyle 2\sin2x\sin^2x=0
\displaystyle \therefore \sin2x=0\quad\text{or}\quad\sin x=0
\displaystyle \sin2x=0\Rightarrow 2x=n\pi\Rightarrow x=\frac{n\pi}{2}
\displaystyle \sin x=0\Rightarrow x=n\pi
\displaystyle \text{For odd }n,\ x=\frac{n\pi}{2}\text{ is inadmissible, since }\tan x\text{ and }\tan3x\text{ are undefined.}
\displaystyle \text{For even }n,\ x=\frac{n\pi}{2}=k\pi.
\displaystyle \therefore x=n\pi,\quad n\in\mathbb Z

\displaystyle \textbf{Question 6: }\text{Solve the following equations:}
\displaystyle \text{i) }\sin x+\cos x=\sqrt{2}\hspace{1cm}\text{ii) }\sqrt{3}\cos x+\sin x=1
\displaystyle \text{iii) }\sin x+\cos x=1\hspace{1cm}\text{iv) }\mathrm{cosec}\,x=1+\cot x
\displaystyle \text{v) }(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2
\displaystyle \text{Answer:}
\displaystyle \text{i) }\sin x+\cos x=\sqrt{2}
\displaystyle \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}
\displaystyle \sin\left(x+\frac{\pi}{4}\right)=1
\displaystyle x+\frac{\pi}{4}=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{\pi}{4}+2n\pi,\quad n\in\mathbb Z

\displaystyle \text{ii) }\sqrt{3}\cos x+\sin x=1
\displaystyle 2\cos\left(x-\frac{\pi}{6}\right)=1
\displaystyle \cos\left(x-\frac{\pi}{6}\right)=\frac{1}{2}
\displaystyle x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{\pi}{2}+2n\pi\quad\text{or}\quad x=-\frac{\pi}{6}+2n\pi,\quad n\in\mathbb Z

\displaystyle \text{iii) }\sin x+\cos x=1
\displaystyle \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=1
\displaystyle \sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}
\displaystyle x+\frac{\pi}{4}=\frac{\pi}{4}+2n\pi\quad\text{or}\quad x+\frac{\pi}{4}=\frac{3\pi}{4}+2n\pi
\displaystyle \therefore x=2n\pi\quad\text{or}\quad x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z

\displaystyle \text{iv) }\mathrm{cosec}\,x=1+\cot x
\displaystyle \frac{1}{\sin x}=1+\frac{\cos x}{\sin x}
\displaystyle \text{Since }\sin x\ne0,\text{ multiplying throughout by }\sin x,
\displaystyle 1=\sin x+\cos x
\displaystyle \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=1
\displaystyle \sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}
\displaystyle x=2n\pi\quad\text{or}\quad x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z
\displaystyle x=2n\pi\text{ is inadmissible, since }\sin x=0.
\displaystyle \therefore x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z

\displaystyle \text{v) }(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2
\displaystyle \text{Now, }\sqrt{3}-1=2\sqrt{2}\sin\frac{\pi}{12}
\displaystyle \text{and }\sqrt{3}+1=2\sqrt{2}\cos\frac{\pi}{12}.
\displaystyle \therefore 2\sqrt{2}\left(\sin\frac{\pi}{12}\cos x+\cos\frac{\pi}{12}\sin x\right)=2
\displaystyle 2\sqrt{2}\sin\left(x+\frac{\pi}{12}\right)=2
\displaystyle \sin\left(x+\frac{\pi}{12}\right)=\frac{1}{\sqrt{2}}
\displaystyle x+\frac{\pi}{12}=\frac{\pi}{4}+2n\pi\quad\text{or}\quad x+\frac{\pi}{12}=\frac{3\pi}{4}+2n\pi
\displaystyle \therefore x=\frac{\pi}{6}+2n\pi\quad\text{or}\quad x=\frac{2\pi}{3}+2n\pi,\quad n\in\mathbb Z

\displaystyle \textbf{Question 7: }\text{Solve the following equations:}
\displaystyle \text{i) }\cot x+\tan x=2\hspace{1cm}\text{ii) }2\sin^2x=3\cos x,\ 0\leq x\leq2\pi
\displaystyle \text{iii) }\sec x\cos5x+1=0,\ 0<x<\frac{\pi}{2}
\displaystyle \text{iv) }5\cos^2x+7\sin^2x-6=0
\displaystyle \text{v) }\sin x-3\sin2x+\sin3x=\cos x-3\cos2x+\cos3x
\displaystyle \text{vi) }4\sin x\cos x+2\sin x+2\cos x+1=0
\displaystyle \text{vii) }\cos x+\sin x=\cos2x+\sin2x
\displaystyle \text{viii) }\sin x\tan x-1=\tan x-\sin x
\displaystyle \text{ix) }3\tan x+\cot x=5\mathrm{cosec}\,x
\displaystyle \text{Answer:}

\displaystyle \text{i) }\cot x+\tan x=2
\displaystyle \frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}=2
\displaystyle \frac{\cos^2x+\sin^2x}{\sin x\cos x}=2
\displaystyle \frac{1}{\sin x\cos x}=2
\displaystyle \sin x\cos x=\frac{1}{2}
\displaystyle \frac{1}{2}\sin2x=\frac{1}{2}
\displaystyle \sin2x=1
\displaystyle 2x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{\pi}{4}+n\pi,\quad n\in\mathbb Z

\displaystyle \text{ii) }2\sin^2x=3\cos x,\quad 0\leq x\leq2\pi
\displaystyle 2(1-\cos^2x)=3\cos x
\displaystyle 2-2\cos^2x=3\cos x
\displaystyle 2\cos^2x+3\cos x-2=0
\displaystyle (2\cos x-1)(\cos x+2)=0
\displaystyle \therefore 2\cos x-1=0\quad\text{or}\quad\cos x+2=0
\displaystyle \cos x=\frac{1}{2}\quad\text{or}\quad\cos x=-2
\displaystyle \cos x=-2\text{ is not possible, since }-1\leq\cos x\leq1.
\displaystyle \therefore \cos x=\frac{1}{2}
\displaystyle \text{For }0\leq x\leq2\pi,\quad x=\frac{\pi}{3},\frac{5\pi}{3}
\displaystyle \therefore x=\frac{\pi}{3}\quad\text{or}\quad x=\frac{5\pi}{3}

\displaystyle \text{iii) }\sec x\cos5x+1=0,\quad 0<x<\frac{\pi}{2}
\displaystyle \frac{\cos5x}{\cos x}+1=0
\displaystyle \text{Since }0<x<\frac{\pi}{2},\text{ we have }\cos x\ne0.
\displaystyle \therefore \cos5x+\cos x=0
\displaystyle \text{Using }\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle 2\cos3x\cos2x=0
\displaystyle \therefore \cos3x=0\quad\text{or}\quad\cos2x=0
\displaystyle \text{Case I: }\cos3x=0
\displaystyle 3x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z
\displaystyle x=\frac{(2n+1)\pi}{6}
\displaystyle \text{For }0<x<\frac{\pi}{2},\quad x=\frac{\pi}{6}.
\displaystyle \text{Case II: }\cos2x=0
\displaystyle 2x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z
\displaystyle x=\frac{(2n+1)\pi}{4}
\displaystyle \text{For }0<x<\frac{\pi}{2},\quad x=\frac{\pi}{4}.
\displaystyle \therefore x=\frac{\pi}{6}\quad\text{or}\quad x=\frac{\pi}{4}

\displaystyle \text{iv) }5\cos^2x+7\sin^2x-6=0
\displaystyle 5(1-\sin^2x)+7\sin^2x-6=0
\displaystyle 5-5\sin^2x+7\sin^2x-6=0
\displaystyle 2\sin^2x-1=0
\displaystyle \sin^2x=\frac{1}{2}
\displaystyle \sin x=\pm\frac{1}{\sqrt{2}}
\displaystyle \therefore x=n\pi\pm\frac{\pi}{4},\quad n\in\mathbb Z

\displaystyle \text{v) }\sin x-3\sin2x+\sin3x=\cos x-3\cos2x+\cos3x
\displaystyle (\sin x+\sin3x)-3\sin2x=(\cos x+\cos3x)-3\cos2x
\displaystyle \text{Using }\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle \sin x+\sin3x=2\sin2x\cos x
\displaystyle \text{Also, using }\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle \cos x+\cos3x=2\cos2x\cos x
\displaystyle \therefore 2\sin2x\cos x-3\sin2x=2\cos2x\cos x-3\cos2x
\displaystyle \sin2x(2\cos x-3)=\cos2x(2\cos x-3)
\displaystyle (2\cos x-3)(\sin2x-\cos2x)=0
\displaystyle \therefore 2\cos x-3=0\quad\text{or}\quad\sin2x-\cos2x=0
\displaystyle \text{Case I: }2\cos x-3=0
\displaystyle \cos x=\frac{3}{2}
\displaystyle \text{This is not possible, since }-1\leq\cos x\leq1.
\displaystyle \text{Case II: }\sin2x-\cos2x=0
\displaystyle \sin2x=\cos2x
\displaystyle \tan2x=1
\displaystyle 2x=n\pi+\frac{\pi}{4},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{n\pi}{2}+\frac{\pi}{8},\quad n\in\mathbb Z

\displaystyle \text{vi) }4\sin x\cos x+2\sin x+2\cos x+1=0
\displaystyle (2\sin x+1)(2\cos x+1)=0
\displaystyle \therefore 2\sin x+1=0\quad\text{or}\quad 2\cos x+1=0
\displaystyle \text{Case I: }2\sin x+1=0
\displaystyle \sin x=-\frac{1}{2}
\displaystyle \therefore x=2n\pi-\frac{\pi}{6}\quad\text{or}\quad x=2n\pi+\frac{7\pi}{6},\quad n\in\mathbb Z
\displaystyle \text{Case II: }2\cos x+1=0
\displaystyle \cos x=-\frac{1}{2}
\displaystyle \therefore x=2n\pi\pm\frac{2\pi}{3},\quad n\in\mathbb Z
\displaystyle \therefore x=2n\pi-\frac{\pi}{6},\quad 2n\pi+\frac{7\pi}{6},\quad\text{or}\quad 2n\pi\pm\frac{2\pi}{3}
\displaystyle \text{where }n\in\mathbb Z.

\displaystyle \text{vii) }\cos x+\sin x=\cos2x+\sin2x
\displaystyle \cos x-\cos2x+\sin x-\sin2x=0
\displaystyle \text{Using }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \cos x-\cos2x=2\sin\frac{3x}{2}\sin\frac{x}{2}
\displaystyle \text{Also, using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \sin x-\sin2x=-2\cos\frac{3x}{2}\sin\frac{x}{2}
\displaystyle \therefore 2\sin\frac{x}{2}\left(\sin\frac{3x}{2}-\cos\frac{3x}{2}\right)=0
\displaystyle \therefore \sin\frac{x}{2}=0\quad\text{or}\quad\sin\frac{3x}{2}=\cos\frac{3x}{2}
\displaystyle \text{Case I: }\sin\frac{x}{2}=0
\displaystyle \frac{x}{2}=n\pi
\displaystyle \therefore x=2n\pi,\quad n\in\mathbb Z
\displaystyle \text{Case II: }\sin\frac{3x}{2}=\cos\frac{3x}{2}
\displaystyle \tan\frac{3x}{2}=1
\displaystyle \frac{3x}{2}=n\pi+\frac{\pi}{4},\quad n\in\mathbb Z
\displaystyle \therefore x=\frac{2n\pi}{3}+\frac{\pi}{6},\quad n\in\mathbb Z
\displaystyle \therefore x=2n\pi\quad\text{or}\quad x=\frac{2n\pi}{3}+\frac{\pi}{6},\quad n\in\mathbb Z

\displaystyle \text{viii) }\sin x\tan x-1=\tan x-\sin x
\displaystyle \frac{\sin^2x}{\cos x}-1=\frac{\sin x}{\cos x}-\sin x
\displaystyle \text{Since }\tan x\text{ is defined, }\cos x\ne0.
\displaystyle \text{Multiplying throughout by }\cos x,
\displaystyle \sin^2x-\cos x=\sin x-\sin x\cos x
\displaystyle \sin^2x+\sin x\cos x-\sin x-\cos x=0
\displaystyle (\sin x-1)(\sin x+\cos x)=0
\displaystyle \therefore \sin x-1=0\quad\text{or}\quad\sin x+\cos x=0
\displaystyle \text{Case I: }\sin x=1
\displaystyle x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z
\displaystyle \text{These values are inadmissible, since }\cos x=0\text{ and }\tan x\text{ is undefined.}
\displaystyle \text{Case II: }\sin x+\cos x=0
\displaystyle \tan x=-1
\displaystyle \therefore x=n\pi-\frac{\pi}{4},\quad n\in\mathbb Z

\displaystyle \text{ix) }3\tan x+\cot x=5\mathrm{cosec}\,x
\displaystyle 3\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{5}{\sin x}
\displaystyle \text{Since the given functions are defined, }\sin x\ne0\text{ and }\cos x\ne0.
\displaystyle \text{Multiplying throughout by }\sin x\cos x,
\displaystyle 3\sin^2x+\cos^2x=5\cos x
\displaystyle 3(1-\cos^2x)+\cos^2x=5\cos x
\displaystyle 3-2\cos^2x=5\cos x
\displaystyle 2\cos^2x+5\cos x-3=0
\displaystyle (2\cos x-1)(\cos x+3)=0
\displaystyle \therefore 2\cos x-1=0\quad\text{or}\quad\cos x+3=0
\displaystyle \cos x=\frac{1}{2}\quad\text{or}\quad\cos x=-3
\displaystyle \cos x=-3\text{ is not possible, since }-1\leq\cos x\leq1.
\displaystyle \therefore \cos x=\frac{1}{2}
\displaystyle \therefore x=2n\pi\pm\frac{\pi}{3},\quad n\in\mathbb Z

\displaystyle \textbf{Question 8: }3-2\cos x-4\sin x-\cos2x+\sin2x=0
\displaystyle \text{Answer:}
\displaystyle 3-2\cos x-4\sin x-\cos2x+\sin2x=0
\displaystyle \Rightarrow 3-\cos2x-4\sin x+\sin2x-2\cos x=0
\displaystyle \Rightarrow 3-(1-2\sin^2x)-4\sin x+2\sin x\cos x-2\cos x=0
\displaystyle \Rightarrow 2\sin^2x+2-4\sin x+2\cos x(\sin x-1)=0
\displaystyle \Rightarrow 2(\sin x-1)^2+2\cos x(\sin x-1)=0
\displaystyle \Rightarrow 2(\sin x-1)(\sin x+\cos x-1)=0
\displaystyle \therefore \sin x-1=0\quad\text{or}\quad\sin x+\cos x-1=0
\displaystyle \text{Case I: }\sin x-1=0
\displaystyle \Rightarrow \sin x=1
\displaystyle \therefore x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z
\displaystyle \text{Case II: }\sin x+\cos x=1
\displaystyle \Rightarrow \frac{1}{\sqrt2}\sin x+\frac{1}{\sqrt2}\cos x=\frac{1}{\sqrt2}
\displaystyle \Rightarrow \sin x\sin\frac{\pi}{4}+\cos x\cos\frac{\pi}{4}=\cos\frac{\pi}{4}
\displaystyle \Rightarrow \cos\left(x-\frac{\pi}{4}\right)=\cos\frac{\pi}{4}
\displaystyle \Rightarrow x-\frac{\pi}{4}=2n\pi\pm\frac{\pi}{4}
\displaystyle \Rightarrow x=2n\pi+\frac{\pi}{2}\quad\text{or}\quad x=2n\pi
\displaystyle \therefore x=2n\pi\quad\text{or}\quad x=\frac{\pi}{2}+2n\pi,\quad n\in\mathbb Z

\displaystyle \textbf{Question 9: }3\sin^2x-5\sin x\cos x+8\cos^2x=2
\displaystyle \text{Answer:}
\displaystyle 3\sin^2x-5\sin x\cos x+8\cos^2x=2
\displaystyle \text{If }\cos x=0,\text{ then }\sin^2x=1.
\displaystyle \text{The left-hand side becomes }3,\text{ which is not equal to }2.
\displaystyle \therefore \cos x\ne0.
\displaystyle \text{Dividing throughout by }\cos^2x,
\displaystyle 3\tan^2x-5\tan x+8=2\sec^2x
\displaystyle \Rightarrow 3\tan^2x-5\tan x+8=2(1+\tan^2x)
\displaystyle \Rightarrow 3\tan^2x-5\tan x+8=2+2\tan^2x
\displaystyle \Rightarrow \tan^2x-5\tan x+6=0
\displaystyle \Rightarrow (\tan x-2)(\tan x-3)=0
\displaystyle \therefore \tan x=2\quad\text{or}\quad\tan x=3
\displaystyle \text{If }\tan x=2,
\displaystyle x=n\pi+\tan^{-1}2,\quad n\in\mathbb Z
\displaystyle \text{If }\tan x=3,
\displaystyle x=n\pi+\tan^{-1}3,\quad n\in\mathbb Z
\displaystyle \therefore x=n\pi+\tan^{-1}2\quad\text{or}\quad x=n\pi+\tan^{-1}3,\quad n\in\mathbb Z

\displaystyle \textbf{Question 10: }2^{\sin^2x}+2^{\cos^2x}=2\sqrt{2}
\displaystyle \text{Answer:}
\displaystyle 2^{\sin^2x}+2^{\cos^2x}=2\sqrt{2}
\displaystyle \Rightarrow 2^{\sin^2x}+2^{1-\sin^2x}=2\sqrt{2}
\displaystyle \text{Let }2^{\sin^2x}=y.
\displaystyle \Rightarrow y+\frac{2}{y}=2\sqrt{2}
\displaystyle \Rightarrow y^2-2\sqrt{2}\,y+2=0
\displaystyle \Rightarrow (y-\sqrt{2})^2=0
\displaystyle \Rightarrow y=\sqrt{2}
\displaystyle \Rightarrow 2^{\sin^2x}=2^{\frac12}
\displaystyle \Rightarrow \sin^2x=\frac12
\displaystyle \Rightarrow \sin x=\pm\frac{1}{\sqrt2}
\displaystyle \Rightarrow \sin x=\pm\sin\frac{\pi}{4}
\displaystyle \therefore x=n\pi\pm\frac{\pi}{4},\quad n\in\mathbb Z

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