Trigonometric Equations

Treating the equation as a quadratic in the single variable $\sin x$ is the cleanest JEE Advanced method, because whenever a trigonometric equation contains only one ratio raised to powers it reduces to an algebraic polynomial. The substitution $t=\sin x$ converts the problem into $2t^2-3t+1=0$, whose factorization $(2t-1)(t-1)=0$ is straightforward. This yields the two simple equations $\sin x=\frac{1}{2}$ and $\sin x=1$, and the periodicity of sine then determines how many angles in the closed interval realize each value. On $[0,2\pi ]$, the value $\sin x=\frac{1}{2}$ is attained at the standard angles $x=\frac{\pi }{6}$ and its supplement $x=\frac{5\pi }{6}$, contributing two solutions. The value $\sin x=1$ is the peak of the sine curve and is attained only once, at $x=\frac{\pi }{2}$, adding a single solution. Summing these gives a total of $2+1=3$ distinct roots in the interval. Option 2 forgets the $\sin x=1$ root entirely. Option 4 wrongly treats the maximum value $\sin x=1$ as if it had two preimages like a generic value. Option 5 over-counts by including the endpoints. This follows the standard quadratic-in-trigonometric-function pattern emphasized throughout JEE Advanced. As a final boundary check, $x=0$ and $x=2\pi$ give $\sin x=0$, which is not a root, so neither endpoint contributes and the count of three stands.