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JEE Main 2026Mathematics Trigonometry

hard
mcq
2026
Official previous-year question

Verified 30 May 2026.

Question

The number of solutions of $2\sin^2 x + \sin^2 2x = 2$ in $[0, 2\pi]$ is:

Options

  1. A

    $6$

  2. B

    $4$

  3. C

    $8$

  4. D

    $2$

Solution

$2\sin^2 x + 4\sin^2 x\cos^2 x = 2$

$\sin^2 x(1 + 2\cos^2 x) = 1$

Let $t = \sin^2 x$: $t(3-2t) = 1 \Rightarrow 2t^2 - 3t + 1 = 0$

$(2t-1)(t-1) = 0 \Rightarrow t = \frac{1}{2}$ or $t = 1$

$\sin^2 x = \frac{1}{2}$: 4 solutions; $\sin^2 x = 1$: 2 solutions. Total $= 6$

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