Class 11: Trigonometric Equations – Multiple Choice Questions

$\displaystyle \textbf{Question 1. }\text{The smallest value of }x\text{ satisfying the equation } \\ \sqrt{3}(\cot x+\tan x)=4\text{ is}$
$\displaystyle \text{(a) }\frac{2\pi}{3}\qquad \text{(b) }\frac{\pi}{3}\qquad \text{(c) }\frac{\pi}{6}\qquad \text{(d) }\frac{\pi}{12}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{3}(\cot x+\tan x)=4$
$\displaystyle \sqrt{3}\left(\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right)=4$
$\displaystyle \sqrt{3}\left(\frac{1}{\sin x\cos x}\right)=4$
$\displaystyle \sin x\cos x=\frac{\sqrt{3}}{4}$
$\displaystyle \frac{1}{2}\sin2x=\frac{\sqrt{3}}{4}$
$\displaystyle \sin2x=\frac{\sqrt{3}}{2}$
$\displaystyle 2x=\frac{\pi}{3}\qquad \text{or}\qquad \frac{2\pi}{3}$
$\displaystyle x=\frac{\pi}{6}\qquad \text{or}\qquad \frac{\pi}{3}$
$\displaystyle \therefore\text{Smallest value of }x=\frac{\pi}{6}$
$\displaystyle \therefore\text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 2. }\text{If }\cos x+\sqrt{3}\sin x=2,\text{ then }x=$
$\displaystyle \text{(a) }\frac{\pi}{3}\qquad \text{(b) }\frac{2\pi}{3}\qquad \text{(c) }\frac{4\pi}{3}\qquad \text{(d) }\frac{5\pi}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \cos x+\sqrt{3}\sin x=2$
$\displaystyle 2\left(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)=2$
$\displaystyle \cos x\cos\frac{\pi}{3}+\sin x\sin\frac{\pi}{3}=1$
$\displaystyle \cos\left(x-\frac{\pi}{3}\right)=1$
$\displaystyle x-\frac{\pi}{3}=0$
$\displaystyle x=\frac{\pi}{3}$
$\displaystyle \therefore\text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 3. }\text{If }\tan px-\tan qx=0,\text{ then the values of }x\text{ form a series in}$
$\displaystyle \text{(a) AP}\qquad \text{(b) GP}\qquad \text{(c) HP}\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \tan px-\tan qx=0$
$\displaystyle \tan px=\tan qx$
$\displaystyle px=qx+n\pi$
$\displaystyle (p-q)x=n\pi$
$\displaystyle x=\frac{n\pi}{p-q},\quad n\in Z$
$\displaystyle \text{The values of }x\text{ form an arithmetic progression.}$
$\displaystyle \therefore\text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 4. }\text{If }a\text{ is any real number, the number of roots of } \\ \cot x-\tan x=a\text{ in the first quadrant is (are).}$
$\displaystyle \text{(a) }2\qquad \text{(b) }0\qquad \text{(c) }1\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \cot x-\tan x=a$
$\displaystyle \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=a$
$\displaystyle \frac{\cos^2x-\sin^2x}{\sin x\cos x}=a$
$\displaystyle \frac{\cos2x}{\frac{1}{2}\sin2x}=a$
$\displaystyle 2\cot2x=a$
$\displaystyle \cot2x=\frac{a}{2}$
$\displaystyle \text{In the first quadrant, }0<x<\frac{\pi}{2}$
$\displaystyle \therefore 0<2x<\pi$
$\displaystyle \text{The equation }\cot2x=\frac{a}{2}\text{ has exactly one solution in }(0,\pi)$
$\displaystyle \therefore\text{Number of roots}=1$
$\displaystyle \therefore\text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 5. }\text{The general solution of the equation }7\cos^2x+3\sin^2x=4\text{ is}$
$\displaystyle \text{(a) }x=2n\pi\pm\frac{\pi}{6},\ n\in Z$
$\displaystyle \text{(b) }x=2n\pi\pm\frac{2\pi}{3},\ n\in Z$
$\displaystyle \text{(c) }x=n\pi\pm\frac{\pi}{3},\ n\in Z$
$\displaystyle \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 7\cos^2x+3\sin^2x=4$
$\displaystyle 7(1-\sin^2x)+3\sin^2x=4$
$\displaystyle 7-4\sin^2x=4$
$\displaystyle \sin^2x=\frac{3}{4}$
$\displaystyle \sin x=\pm\frac{\sqrt{3}}{2}$
$\displaystyle x=n\pi\pm\frac{\pi}{3},\ n\in Z$
$\displaystyle \therefore\text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 6. }\text{A solution of the equation }\cos^2x+\sin x+1=0,\text{ lies in the interval}$
$\displaystyle \text{(a) }\left(-\frac{\pi}{4},\frac{\pi}{4}\right)\qquad \text{(b) }\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$
$\displaystyle \text{(c) }\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\qquad \text{(d) }\left(\frac{5\pi}{4},\frac{7\pi}{4}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \cos^2x+\sin x+1=0$
$\displaystyle 1-\sin^2x+\sin x+1=0$
$\displaystyle \sin^2x-\sin x-2=0$
$\displaystyle (\sin x-2)(\sin x+1)=0$
$\displaystyle \sin x=-1$
$\displaystyle x=\frac{3\pi}{2}$
$\displaystyle \frac{3\pi}{2}\in\left(\frac{5\pi}{4},\frac{7\pi}{4}\right)$
$\displaystyle \therefore\text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 7. }\text{The number of solution in }\left[0,\frac{\pi}{2}\right]\text{ of the equation } \\ \cos3x\tan5x=\sin7x\text{ is}$
$\displaystyle \text{(a) }5\qquad \text{(b) }7\qquad \text{(c) }6\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \cos3x\tan5x=\sin7x$
$\displaystyle \cos3x\cdot\frac{\sin5x}{\cos5x}=\sin7x$
$\displaystyle \cos3x\sin5x=\sin7x\cos5x$
$\displaystyle \sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$
$\displaystyle \frac{1}{2}[\sin8x+\sin2x]=\frac{1}{2}[\sin12x+\sin2x]$
$\displaystyle \sin8x=\sin12x$
$\displaystyle 8x=12x+2n\pi\qquad \text{or}\qquad 8x=\pi-12x+2n\pi$
$\displaystyle x=-\frac{n\pi}{2}\qquad \text{or}\qquad x=\frac{(2n+1)\pi}{20}$
$\displaystyle \text{From }x=-\frac{n\pi}{2},\text{ only }x=0\text{ lies in }\left[0,\frac{\pi}{2}\right]$
$\displaystyle \text{From }x=\frac{(2n+1)\pi}{20},\text{ we get }x=\frac{\pi}{20},\frac{3\pi}{20},\frac{5\pi}{20},\frac{7\pi}{20},\frac{9\pi}{20}$
$\displaystyle \text{But at }x=\frac{\pi}{10},\frac{3\pi}{10},\frac{\pi}{2},\ \tan5x\text{ is not defined}$
$\displaystyle \therefore \text{Valid solutions are }x=0,\frac{\pi}{20},\frac{3\pi}{20},\frac{7\pi}{20},\frac{9\pi}{20}$
$\displaystyle \therefore \text{Number of solutions}=5$
$\displaystyle \therefore\text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 8. }\text{The general value of }x\text{ satisfying the equation } \\ \sqrt{3}\sin x+\cos x=\sqrt{3}\text{ is given by}$
$\displaystyle \text{(a) }x=n\pi+(-1)^n\frac{\pi}{4}+\frac{\pi}{3},\ n\in Z$
$\displaystyle \text{(b) }x=n\pi+(-1)^n\frac{\pi}{3}-\frac{\pi}{6},\ n\in Z$
$\displaystyle \text{(c) }x=n\pi\pm\frac{\pi}{6},\ n\in Z$
$\displaystyle \text{(d) }x=n\pi\pm\frac{\pi}{3},\ n\in Z$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{3}\sin x+\cos x=\sqrt{3}$
$\displaystyle 2\left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right)=\sqrt{3}$
$\displaystyle 2\sin\left(x+\frac{\pi}{6}\right)=\sqrt{3}$
$\displaystyle \sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$
$\displaystyle x+\frac{\pi}{6}=\frac{\pi}{3}+2n\pi\qquad \text{or}\qquad \frac{2\pi}{3}+2n\pi$
$\displaystyle x=\frac{\pi}{6}+2n\pi\qquad \text{or}\qquad \frac{\pi}{2}+2n\pi$
$\displaystyle \therefore x=n\pi+(-1)^n\frac{\pi}{6}+\frac{\pi}{3},\ n\in Z$
$\displaystyle \therefore\text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 9. }\text{The smallest positive angle which satisfies the equation } \\ 2\sin^2x+\sqrt{3}\cos x+1=0\text{ is}$
$\displaystyle \text{(a) }\frac{5\pi}{6}\qquad \text{(b) }\frac{2\pi}{3}\qquad \text{(c) }\frac{\pi}{3}\qquad \text{(d) }\frac{\pi}{6}$
$\displaystyle \text{Answer:}$
$\displaystyle 2\sin^2x+\sqrt{3}\cos x+1=0$
$\displaystyle 2(1-\cos^2x)+\sqrt{3}\cos x+1=0$
$\displaystyle 2\cos^2x-\sqrt{3}\cos x-3=0$
$\displaystyle (2\cos x+\sqrt{3})(\cos x-\sqrt{3})=0$
$\displaystyle \cos x=-\frac{\sqrt{3}}{2}$
$\displaystyle \therefore x=\frac{5\pi}{6},\frac{7\pi}{6}$
$\displaystyle \therefore\text{Smallest positive angle}=\frac{5\pi}{6}$
$\displaystyle \therefore\text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 10. }\text{If }4\sin^2x=1,\text{ then the values of }x\text{ are}$
$\displaystyle \text{(a) }2n\pi\pm\frac{\pi}{3},\ n\in Z\qquad \text{(b) }n\pi\pm\frac{\pi}{3},\ n\in Z$
$\displaystyle \text{(c) }n\pi+\frac{\pi}{6},\ n\in Z\qquad \text{(d) }2n\pi\pm\frac{\pi}{6},\ n\in Z$
$\displaystyle \text{Answer:}$
$\displaystyle 4\sin^2x=1$
$\displaystyle \sin^2x=\frac{1}{4}$
$\displaystyle \sin x=\pm\frac{1}{2}$
$\displaystyle \therefore x=n\pi\pm\frac{\pi}{6},\ n\in Z$
$\displaystyle \therefore\text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 11. }\text{If }\cot x-\tan x=\sec x,\text{ then }x\text{ is equal to}$
$\displaystyle \text{(a) }2n\pi+\frac{3\pi}{2},\ n\in Z\qquad \text{(b) }n\pi+(-1)^n\frac{\pi}{6},\ n\in Z$
$\displaystyle \text{(c) }n\pi+\frac{\pi}{2},\ n\in Z\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \cot x-\tan x=\sec x$
$\displaystyle \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=\frac{1}{\cos x}$
$\displaystyle \frac{\cos^2x-\sin^2x}{\sin x\cos x}=\frac{1}{\cos x}$
$\displaystyle \frac{\cos2x}{\sin x\cos x}=\frac{1}{\cos x}$
$\displaystyle \cos2x=\sin x$
$\displaystyle 1-2\sin^2x=\sin x$
$\displaystyle 2\sin^2x+\sin x-1=0$
$\displaystyle (2\sin x-1)(\sin x+1)=0$
$\displaystyle \sin x=\frac{1}{2}\qquad \text{or}\qquad \sin x=-1$
$\displaystyle \sin x=-1\text{ is not valid because }\cos x=0$
$\displaystyle \therefore \sin x=\frac{1}{2}$
$\displaystyle x=n\pi+(-1)^n\frac{\pi}{6},\ n\in Z$
$\displaystyle \therefore\text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 12. }\text{A value of }x\text{ satisfying }\cos x+\sqrt{3}\sin x=2\text{ is}$
$\displaystyle \text{(a) }\frac{5\pi}{3}\qquad \text{(b) }\frac{4\pi}{3}\qquad \text{(c) }\frac{2\pi}{3}\qquad \text{(d) }\frac{\pi}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \cos x+\sqrt{3}\sin x=2$
$\displaystyle 2\left(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)=2$
$\displaystyle \cos x\cos\frac{\pi}{3}+\sin x\sin\frac{\pi}{3}=1$
$\displaystyle \cos\left(x-\frac{\pi}{3}\right)=1$
$\displaystyle x-\frac{\pi}{3}=2n\pi$
$\displaystyle x=2n\pi+\frac{\pi}{3}$
$\displaystyle \therefore\text{A value of }x=\frac{\pi}{3}$
$\displaystyle \therefore\text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 13. }\text{In }(0,\pi),\text{ the number of solutions of the equation } \\ \tan x+\tan2x+\tan3x=\tan x\tan2x\tan3x\text{ is}$
$\displaystyle \text{(a) }7\qquad \text{(b) }5\qquad \text{(c) }4\qquad \text{(d) }2$
$\displaystyle \text{Answer:}$
$\displaystyle \tan x+\tan2x+\tan3x=\tan x\tan2x\tan3x$
$\displaystyle \text{Using }\tan(A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}$
$\displaystyle \tan(x+2x+3x)=0$
$\displaystyle \tan6x=0$
$\displaystyle 6x=n\pi$
$\displaystyle x=\frac{n\pi}{6}$
$\displaystyle \text{In }(0,\pi),\ n=1,2,3,4,5$
$\displaystyle \therefore x=\frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6}$
$\displaystyle \text{But at }x=\frac{\pi}{2},\ \tan x\text{ is not defined}$
$\displaystyle \therefore \text{Valid solutions are }\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6}$
$\displaystyle \therefore \text{Number of solutions}=4$
$\displaystyle \therefore\text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 14. }\text{The number of values of }x\text{ in }[0,2\pi]\text{ that satisfy the equation } \\ \sin^2x-\cos x=\frac{1}{4}\text{ is}$
$\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }3\qquad \text{(d) }4$
$\displaystyle \text{Answer:}$
$\displaystyle \sin^2x-\cos x=\frac{1}{4}$
$\displaystyle 1-\cos^2x-\cos x=\frac{1}{4}$
$\displaystyle \cos^2x+\cos x-\frac{3}{4}=0$
$\displaystyle 4\cos^2x+4\cos x-3=0$
$\displaystyle (2\cos x-1)(2\cos x+3)=0$
$\displaystyle \cos x=\frac{1}{2}\qquad \text{or}\qquad \cos x=-\frac{3}{2}$
$\displaystyle \cos x=-\frac{3}{2}\text{ is not possible}$
$\displaystyle \therefore \cos x=\frac{1}{2}$
$\displaystyle x=\frac{\pi}{3},\frac{5\pi}{3}$
$\displaystyle \therefore\text{Number of values}=2$
$\displaystyle \therefore\text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 15. }\text{If }e^{\sin x}-e^{-\sin x}-4=0,\text{ then }x=$
$\displaystyle \text{(a) }0\qquad \text{(b) }\sin^{-1}\{\log_e(2-\sqrt{5})\}\qquad \text{(c) }1\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle e^{\sin x}-e^{-\sin x}-4=0$
$\displaystyle \text{Let }e^{\sin x}=y$
$\displaystyle y-\frac{1}{y}-4=0$
$\displaystyle y^2-4y-1=0$
$\displaystyle y=2\pm\sqrt{5}$
$\displaystyle e^{\sin x}=2+\sqrt{5}$
$\displaystyle \sin x=\log_e(2+\sqrt{5})$
$\displaystyle \text{But }\log_e(2+\sqrt{5})>1$
$\displaystyle \therefore\text{No real value of }x\text{ is possible}$
$\displaystyle \therefore\text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 16. }\text{The equation }3\cos x+4\sin x=6\text{ has .... solution}$
$\displaystyle \text{(a) finite}\qquad \text{(b) infinite}\qquad \text{(c) one}\qquad \text{(d) no}$
$\displaystyle \text{Answer:}$
$\displaystyle 3\cos x+4\sin x=6$
$\displaystyle \text{Maximum value of }3\cos x+4\sin x=\sqrt{3^2+4^2}=5$
$\displaystyle \text{But }6>5$
$\displaystyle \therefore\text{The equation has no solution}$
$\displaystyle \therefore\text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 17. }\text{If }\sqrt{3}\cos x+\sin x=\sqrt{2},\text{ then general value of }x\text{ is}$
$\displaystyle \text{(a) }n\pi+(-1)^n\frac{\pi}{4},\ n\in Z$
$\displaystyle \text{(b) }(-1)^n\frac{\pi}{4}-\frac{\pi}{3},\ n\in Z$
$\displaystyle \text{(c) }n\pi+\frac{\pi}{4}-\frac{\pi}{3},\ n\in Z$
$\displaystyle \text{(d) }n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{3},\ n\in Z$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{3}\cos x+\sin x=\sqrt{2}$
$\displaystyle 2\left(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right)=\sqrt{2}$
$\displaystyle 2\cos\left(x-\frac{\pi}{6}\right)=\sqrt{2}$
$\displaystyle \cos\left(x-\frac{\pi}{6}\right)=\frac{1}{\sqrt{2}}$
$\displaystyle x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{4}$
$\displaystyle x=2n\pi+\frac{\pi}{6}\pm\frac{\pi}{4}$
$\displaystyle \therefore x=2n\pi-\frac{\pi}{12}\quad \text{or}\quad x=2n\pi+\frac{5\pi}{12}$
$\displaystyle \therefore x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{3},\ n\in Z$
$\displaystyle \therefore\text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 18. }\text{General solution of }\tan5x=\cot2x\text{ is}$
$\displaystyle \text{(a) }\frac{n\pi}{7}+\frac{\pi}{2},\ n\in Z\qquad \text{(b) }x=\frac{n\pi}{7}+\frac{\pi}{3},\ n\in Z$
$\displaystyle \text{(c) }x=\frac{n\pi}{7}+\frac{\pi}{14},\ n\in Z\qquad \text{(d) }x=\frac{n\pi}{7}-\frac{\pi}{14},\ n\in Z$
$\displaystyle \text{Answer:}$
$\displaystyle \tan5x=\cot2x$
$\displaystyle \tan5x=\tan\left(\frac{\pi}{2}-2x\right)$
$\displaystyle 5x=n\pi+\frac{\pi}{2}-2x$
$\displaystyle 7x=n\pi+\frac{\pi}{2}$
$\displaystyle x=\frac{n\pi}{7}+\frac{\pi}{14},\ n\in Z$
$\displaystyle \therefore\text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 19. }\text{The solution of the equation }\cos^2x+\sin x+1=0\text{ lies in the interval}$
$\displaystyle \text{(a) }\left(-\frac{\pi}{4},\frac{\pi}{4}\right)\qquad \text{(b) }\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$
$\displaystyle \text{(c) }\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\qquad \text{(d) }\left(\frac{5\pi}{4},\frac{7\pi}{4}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \cos^2x+\sin x+1=0$
$\displaystyle 1-\sin^2x+\sin x+1=0$
$\displaystyle \sin^2x-\sin x-2=0$
$\displaystyle (\sin x-2)(\sin x+1)=0$
$\displaystyle \sin x=-1$
$\displaystyle x=\frac{3\pi}{2}$
$\displaystyle \frac{3\pi}{2}\in\left(\frac{5\pi}{4},\frac{7\pi}{4}\right)$
$\displaystyle \therefore\text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 20. }\text{If }\cos x=-\frac{1}{2}\text{ and }0<x<2\pi,\text{ then the solutions are}$
$\displaystyle \text{(a) }x=\frac{\pi}{3},\frac{4\pi}{3}\qquad \text{(b) }x=\frac{2\pi}{3},\frac{4\pi}{3}$
$\displaystyle \text{(c) }x=\frac{2\pi}{3},\frac{7\pi}{6}\qquad \text{(d) }\theta=\frac{2\pi}{3},\frac{5\pi}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \cos x=-\frac{1}{2}$
$\displaystyle \text{In }0<x<2\pi,\text{ cosine is negative in the second and third quadrants}$
$\displaystyle \therefore x=\frac{2\pi}{3},\frac{4\pi}{3}$
$\displaystyle \therefore\text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 21. }\text{The number of values of }x\text{ in the interval }[0,5\pi]\text{ satisfying } \\ 3\sin^2x-7\sin x+2=0\text{ is}$
$\displaystyle \text{(a) }0\qquad \text{(b) }5\qquad \text{(c) }6\qquad \text{(d) }10$
$\displaystyle \text{Answer:}$
$\displaystyle 3\sin^2x-7\sin x+2=0$
$\displaystyle (3\sin x-1)(\sin x-2)=0$
$\displaystyle \sin x=\frac{1}{3}\qquad \text{or}\qquad \sin x=2$
$\displaystyle \sin x=2\text{ is not possible}$
$\displaystyle \therefore \sin x=\frac{1}{3}$
$\displaystyle \text{In each interval of length }2\pi,\text{ there are }2\text{ solutions}$
$\displaystyle \text{In }[0,5\pi],\text{ there are }2+2+2=6\text{ solutions}$
$\displaystyle \therefore\text{Correct option is (c).}$
$\\$