MHTCET 17 Apr 2026 Shift 1 Paper

Concept:Convert the trigonometric equation into a polynomial in $\sin\theta$, then count valid roots in the given interval.Explanation:Let $x = \sin\theta$.Since $\theta \in [0,\pi]$, we have $x \in [0,1]$.Use the identities:$$\sin 3\theta = 3\sin\theta - 4\sin^3\theta$$$$\cos^2\theta = 1 - \sin^2\theta$$Substitute into the given equation:$$3x - 4x^3 - (1 - x^2) = \frac{1}{4}$$Simplify:$$-4x^3 + x^2 + 3x - \frac{5}{4} = 0$$Multiply by $-4$:$$16x^3 - 4x^2 - 12x + 5 = 0$$Factor the cubic:$$(2x - 1)(8x^2 + 2x - 5) = 0$$So, one root is:$$x = \frac{1}{2}$$Solving the quadratic:$$8x^2 + 2x - 5 = 0$$$$x = \frac{-2 \pm \sqrt{4 + 160}}{16}$$$$x = \frac{-1 \pm \sqrt{41}}{8}$$Only $x = \frac{-1 + \sqrt{41}}{8}$ lies in $[0,1]$, and the other root is negative.So the valid values of $\sin\theta$ are:$$\frac{1}{2}, \quad \frac{-1 + \sqrt{41}}{8}$$Both lie in $(0,1)$.For each such value of $\sin\theta$, there are exactly two values of $\theta$ in $[0,\pi]$.Thus, total number of solutions:$$2 \times 2 = 4$$Answer:$$4$$Hence, the correct option is C.