(sin2x+√3 cos2x )^2=5+cos(π/6-2x)

Решите задачу:

$(sin2x+\sqrt3cos2x)^2=5+cos( \frac{\pi}{6}-2x)\\\\\\\star \; \; sin2x+\sqrt3cos2x=2\cdot ( \frac{1}{2}\cdot sin2x+\frac{\sqrt3}{2}\cdot cos2x)=\\\\=2\cdot (sin\frac{\pi}{6}\cdot sin2x+cos \frac{\pi}{6}\cdot cos2x)=2\cdot cos( \frac{\pi}{6} -2x)\; \; \star \\\\\\4cos^2( \frac{\pi}{6}-2x) -cos( \frac{\pi}{6}-2x)-5=0 \\\\t=cos( \frac{\pi}{6} -2x)\; ,\; -1 \leq t \leq 1\; ,\; \;5t^2-t-5=0\; ,\\\\D=81\; ,\; \; t_1=\frac{1-9}{8}=-1\; ,\; t_2=\frac{10}{8}=\frac{5}{4}\ \textgreater \ 1\\\\cos( \frac{\pi}{6}-2x)= \frac{1-9}{8}=-1$

$\frac{\pi}{6} -2x=-\pi +2\pi n,\; n\in Z\\\\2x=-\frac{5\pi}{6}+2\pi n\; ,\; n\in Z\quad (period\; mozno\; pisat:\; \; \pm 2\pi n)\\\\x=-\frac{5\pi}{12}+\pi n,\; n\in Z$