Question

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

Answer 1

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

$1+\frac{\cos 2\theta }{2}+1+\frac{\cos \left(2(x+\frac{\pi }{3})\right)}{2}+1+\frac{\cos \left(2(x-\frac{\pi }{3})\right)}{2}$

$=\frac{1}{2}(1+\cos 2x+1+\cos (2x+\frac{2\pi }{3})+1+\cos (2x-\frac{2\pi }{3}\left)\right)$

$=\frac{1}{2}(3+\cos 2x+\cos (2x+\frac{2\pi }{3})+\cos (2x-\frac{2\pi }{3}))$

$\cos =\frac{1}{2}(3+\cos 2x+2\cos \left(\frac{(2x+\frac{2\pi }{3}+2x-\frac{2\pi }{5})}{2}\right)\cos (2x+\frac{2\pi }{3}-2x+\frac{2\pi }{3}))$

$=\frac{1}{2}(3+\cos 2x+2\cos 2x\cos \frac{2\pi }{3})$

$=\frac{1}{2}(3+\cos 2x+2\cdot \cos 2x\left(\frac{1}{2}\right))$

$=\frac{1}{2}(3+\cos 2x-\cos 2x)$

$=\frac{3}{2}$.