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
Solution
<p>$$\begin{aligned}
& 2 \sin ^2 \theta=\cos 2 \theta \\
& 2 \sin ^2 \theta=1-2 \sin ^2 \theta \\
& 4 \sin ^2 \theta=1 \\
& \sin ^2 \theta=\frac{1}{4} \\
& \sin \theta= \pm \frac{1}{2} \\
& 2 \cos ^2 \theta=3 \sin \theta \\
& 2-2 \sin ^2 \theta+3 \sin \theta-2=0 \\
& (2 \sin \theta-1)(2 \sin \theta-2)=0 \\
& \sin \theta=\frac{1}{2}
\end{aligned}$$</p>
<p>so common equation which satisfy both equations is $\sin \theta=\frac{1}{2}$</p>
<p>$\theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])$</p>
<p>$\text { Sum }=\pi$</p>
About this question
Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Ratios and Identities
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