JEE Main · 2006 · 30 Apr 2006 · Offline · Trigonometric Equations

Treat the equation as a quadratic in $\sin x$: $2{\sin }^{2}x+5\sin x-3=0$ factors as $(2\sin x-1)(\sin x+3)=0$, giving $\sin x=\frac{1}{2}$ or $\sin x=-3$. The second is impossible since $|\sin x|\le 1$, so only $\sin x=\frac{1}{2}$ survives. On $[0,\pi ]$ this is met at $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$; on $[\pi ,2\pi ]$ the sine is negative so there are none; and on $[2\pi ,3\pi ]$ the two solutions repeat at $2\pi +\frac{\pi }{6}$ and $2\pi +\frac{5\pi }{6}$. Counting over $[0,3\pi ]$ gives four values of $x$.