JEE Mathematics 2025 Question with Solution

Given: We need the sum of all values of $\theta \in [0, 2\pi]$ satisfying both $2\sin^2\theta = \cos 2\theta$ and $2\cos^2\theta = 3\sin\theta$.

Find: The sum of all common values of $\theta$ in the given interval.

Using $\cos 2\theta = 1 - 2\sin^2\theta$ in the first equation,

$$2\sin^2\theta = 1 - 2\sin^2\theta$$

so,

$$4\sin^2\theta = 1$$

which gives

$$\sin^2\theta = \frac{1}{4}$$

and hence

$$\sin\theta = \frac{1}{2} \quad \text{or} \quad \sin\theta = -\frac{1}{2}$$

Therefore, from the first equation,

$$\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$

Now use $\cos^2\theta = 1 - \sin^2\theta$ in the second equation:

$$2(1 - \sin^2\theta) = 3\sin\theta$$

so,

$$2 - 2\sin^2\theta = 3\sin\theta$$

Rearranging,

$$2\sin^2\theta + 3\sin\theta - 2 = 0$$

Let $x = \sin\theta$. Then

$$2x^2 + 3x - 2 = 0$$

Solving,

$$x = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$$

Thus,

$$x = \frac{1}{2} \quad \text{or} \quad x = -2$$

Since $\sin\theta$ cannot be $-2$, we get

$$\sin\theta = \frac{1}{2}$$

Therefore, the common solutions are

$$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$$

Their sum is

$$\frac{\pi}{6} + \frac{5\pi}{6} = \pi$$

Therefore, the sum of all solutions is $\pi$. The correct option is C.

Given: Both equations must be satisfied simultaneously.

Find: The common values of $\theta$.

From the first equation,

$$2\sin^2\theta = \cos 2\theta = 1 - 2\sin^2\theta$$

which gives

$$\sin\theta = \pm \frac{1}{2}$$

From the second equation, substituting $\cos^2\theta = 1 - \sin^2\theta$ gives a quadratic in $\sin\theta$:

$$2\sin^2\theta + 3\sin\theta - 2 = 0$$

This factors to admissible value $\sin\theta = \frac{1}{2}$, since the other value $-2$ is not possible for sine.

So the common condition is immediately

$$\sin\theta = \frac{1}{2}$$

Hence in $[0,2\pi]$,

$$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$$

and the sum is $\pi$. The correct option is C.