Question

Find the solution of the equation:

${\cos }^{2}x+{\cos }^{2}2x+{\cos }^{2}3x=1$

Answer 1

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

${\cos }^2\left(x\right)-\frac{1}{2}+{\cos }^2\left(2x\right)+{\cos }^2\left(3x\right)-\frac{1}{2}=0$

$\frac{\cos \left(2x\right)+\cos \left(6x\right)}{2}+{\cos }^2\left(2x\right)=0$

$\cos \left(4x\right)\cos \left(2x\right)+{\cos }^2\left(2x\right)=0$

$\left(\cos \right(4x)+\cos (2x\left)\right)\cos \left(2x\right)=0$

$2\cos \left(3x\right)\cos \left(x\right)\cos \left(2x\right)=0$

$\cos \left(x\right)=0$ implies $\cos \left(3x\right)=0$, so the $\cos \left(x\right)$ factor can be ignored.

Hence $3x=\frac{\pi }{2}+k\pi$ or $2x=\frac{\pi }{2}+k\pi ,$ where $k$ is an integer,

i.e., $x=\frac{\pi }{6}+\frac{k\pi }{3}$ or $\frac{\pi }{4}+\frac{k\pi }{2}$, where $k$ is an integer.