ПОМОГИТЕ3sin2x+4cos2x=5

Решите задачу:

$3sin2x+4cos2x=5\\\\\frac{\sqrt3}{5}}sin2x+\frac{2}{5}cos2x=1\\\\cos \alpha =\frac{\sqrt3}{5}\; ,\; \; sin \alpha =\frac{2}{5}\; \; \to \; \; tg \alpha =\frac{2}{\sqrt3}\; \; \to \; \; \alpha =arctg\frac{2}{\sqrt3}\\\\sin(2x+ \alpha )=1\\\\2x+ \alpha =\frac{\pi}{2}+2\pi n\; ,\; n\in Z\\\\2x=\frac{\pi}{2}- \alpha +2\pi n=\frac{\pi}{2}-arctg\frac{2}{\sqrt3}+2\pi n\; ,\; n\in Z\\\\x=\frac{\pi}{4}-\frac{1}{2}arctg\frac{2}{\sqrt3}+\pi n\; ,\; n\in Z$