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
Step-by-Step Solution
Key Concept: Reduce both equations to conditions on $\sin\theta$ independently, then retain only the values of $\theta$ satisfying both simultaneously.
**Eq. 1:** $2\sin^2\theta = \cos 2\theta = 1-2\sin^2\theta \Rightarrow 4\sin^2\theta = 1 \Rightarrow \sin\theta = \pm\tfrac{1}{2}$.
**Eq. 2:** $2\cos^2\theta = 3\sin\theta \Rightarrow 2(1-\sin^2\theta)=3\sin\theta \Rightarrow 2\sin^2\theta+3\sin\theta-2=0 \Rightarrow (2\sin\theta-1)(\sin\theta+2)=0 \Rightarrow \sin\theta = \tfrac{1}{2}$.
**Common solutions:** $\sin\theta=\tfrac{1}{2}$ gives $\theta=\dfrac{\pi}{6},\,\dfrac{5\pi}{6}$ in $[0,2\pi]$.
**Sum** $= \dfrac{\pi}{6}+\dfrac{5\pi}{6} = \pi$.
Correct Answer: 3