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The principal solutions of $\cos 2 x=\frac{-1}{2}$ are
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The correct answer is:
$x=\frac{\pi}{3}, \quad x=\frac{2 \pi}{3}$
(B)
$\begin{aligned} & \cos 2 x=\frac{-1}{2} \\ \therefore & \frac{-1}{2}=\cos \left(\pi-\frac{\pi}{3}\right)-\cos \left(\pi+\frac{\pi}{3}\right) \Rightarrow \frac{-1}{2}-\cos \frac{2 \pi}{3}-\cos \frac{4 \pi}{3} \\ \therefore & 2 x=\frac{2 \pi}{3} \text { or } 2 x=\frac{4 \pi}{3} \Rightarrow x=\frac{\pi}{3} \text { or } \frac{2 \pi}{3} \end{aligned}$
$\begin{aligned} & \cos 2 x=\frac{-1}{2} \\ \therefore & \frac{-1}{2}=\cos \left(\pi-\frac{\pi}{3}\right)-\cos \left(\pi+\frac{\pi}{3}\right) \Rightarrow \frac{-1}{2}-\cos \frac{2 \pi}{3}-\cos \frac{4 \pi}{3} \\ \therefore & 2 x=\frac{2 \pi}{3} \text { or } 2 x=\frac{4 \pi}{3} \Rightarrow x=\frac{\pi}{3} \text { or } \frac{2 \pi}{3} \end{aligned}$
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