$\cos ^2(x)+\cos ^2\left(x+\frac{\pi}{3}\right)+\cos ^2\left(x-\frac{\pi}{3}\right)=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \cos ^2 x+\cos ^2\left(x+\frac{\pi}{3}\right)+\cos ^2\left(x-\frac{\pi}{3}\right) \ & \because \quad \cos 2 x=2 \cos ^2 x-1 \ & \Rightarrow \quad \cos ^2 x=\frac{\cos 2 x+1}{2} \ & \Rightarrow \quad\left[\frac{1+\cos 2 x}{2}\right]+\left[\frac{1+\cos \left(2 x+\frac{2 \pi}{3}\right)}{2}\right] \ & +\left[\frac{1+\cos \left(2 x-\frac{2 \pi}{3}\right)}{2}\right] \ & \Rightarrow \quad \frac{1}{2}\left[1+\cos 2 x+1+\cos \left(2 x+\frac{2 \pi}{3}\right)\right. \ & \left.+1+\cos \left(2 x-\frac{2 \pi}{3}\right)\right] \ & \Rightarrow \quad \frac{1}{2}\left[3+\cos 2 x+\cos \left(2 x+\frac{2 \pi}{3}\right)\right. \ & \left.+\cos \left(2 x-\frac{2 \pi}{3}\right)\right] \ & \Rightarrow \quad \frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cdot \cos \frac{2 \pi}{3}\right] \ & \Rightarrow \quad \frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x\left(-\frac{1}{2}\right)\right] \ & \Rightarrow \quad \frac{1}{2}[3+\cos 2 x-\cos 2 x]=\left(\frac{3}{2}\right) \ \end{aligned}$

\(\cos ^2(x)+\cos ^2\left(x+\frac{\pi}{3}\right)+\cos ^2\left(x-\frac{\pi}{3}\right)=\)

Solution