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

We have $\text{L.H.S.} = \cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)$
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$\text{L.H.S.} = \frac{1 + \cos 2x}{2} + \frac{1 + \cos(2x + \frac{2\pi}{3})}{2} + \frac{1 + \cos(2x - \frac{2\pi}{3})}{2}$
$= \frac{1}{2} [3 + \cos 2x + \cos(2x + \frac{2\pi}{3}) + \cos(2x - \frac{2\pi}{3})]$
Using the formula $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$:
$= \frac{1}{2} [3 + \cos 2x + 2\cos 2x \cos \frac{2\pi}{3}]$
Since $\cos \frac{2\pi}{3} = -\frac{1}{2}$:
$= \frac{1}{2} [3 + \cos 2x + 2\cos 2x(-\frac{1}{2})]$
$= \frac{1}{2} [3 + \cos 2x - \cos 2x]$
$= \frac{3}{2} = \text{R.H.S.}$