How do you solve the equation 5cos(2x)=2 ?
I would divide through by 5 first. Then I would solve the 2 cases (where $\displaystyle k\in\mathbb{Z}$):
a) $\displaystyle \cos(2x)=\cos(2x+2k\pi)=\frac{2}{5}$
b) $\displaystyle \cos(2x)=\cos(2k\pi-2x)=\frac{2}{5}$
Recall, if $\displaystyle \cos(f(x))=a$ then $\displaystyle f(x)=\cos^{-1}(a)$, where $\displaystyle a$ is an arbitrary constant.