The principal solutions of $\cos 2 x=\frac{-1}{2}$ are

The principal solutions of $\cos 2 x=\frac{-1}{2}$ are
  1. $x=\frac{-2 \pi}{3}, x=\frac{4 \pi}{3}$
  2. $x=\frac{\pi}{3}, \quad x=\frac{2 \pi}{3}$
  3. $x=\frac{-\pi}{3}, \quad x=\frac{5 \pi}{6}$
  4. $x=\frac{\pi}{3}, \quad x=\frac{7 \pi}{6}$

Solution

$\begin{aligned} & \cos 2 x=\frac{-1}{2} \\ \therefore & \frac{-1}{2}=\cos \left(\pi-\frac{\pi}{3}\right)-\cos \left(\pi+\frac{\pi}{3}\right) \Rightarrow \frac{-1}{2}-\cos \frac{2 \pi}{3}-\cos \frac{4 \pi}{3} \\ \therefore & 2 x=\frac{2 \pi}{3} \text { or } 2 x=\frac{4 \pi}{3} \Rightarrow x=\frac{\pi}{3} \text { or } \frac{2 \pi}{3} \end{aligned}$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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