JEE MAIN - Mathematics (2025 - 22nd January Evening Shift - No. 4)
The sum of all values of $\theta \in[0,2 \pi]$ satisfying $2 \sin ^2 \theta=\cos 2 \theta$ and $2 \cos ^2 \theta=3 \sin \theta$ is
$\pi$
$\frac{5 \pi}{6}$
$\frac{\pi}{2}$
$4 \pi$
Explanation
$$\begin{aligned} & 2 \sin ^2 \theta=\cos 2 \theta \\ & 2 \sin ^2 \theta=1-2 \sin ^2 \theta \\ & 4 \sin ^2 \theta=1 \\ & \sin ^2 \theta=\frac{1}{4} \\ & \sin \theta= \pm \frac{1}{2} \\ & 2 \cos ^2 \theta=3 \sin \theta \\ & 2-2 \sin ^2 \theta+3 \sin \theta-2=0 \\ & (2 \sin \theta-1)(2 \sin \theta-2)=0 \\ & \sin \theta=\frac{1}{2} \end{aligned}$$
so common equation which satisfy both equations is $\sin \theta=\frac{1}{2}$
$$\theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])$$
$$\text { Sum }=\pi$$
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