Sin2x+2√3cos^x-6sinx-6√3cosx=0,на промежутке [pi/2;2*pi].

Решите задачу:

$\sin2x+2\sqrt3\cos^2x-6\sin x-6\sqrt3\cos x=0\\\\2\sin x\cos x-6\sin x+2\sqrt3\cos^2x-6\sqrt3\cos x=0\\2\sin x(\cos x-3)+2\sqrt3\cos x(\cos x-3)=0\\(2\sin x+2\sqrt3\cos x)(\cos x-3)=0\\\begin{cases}2\sin x+2\sqrt3\cos x=0\\\cos x-3=0\;-\;pew.\;HET\end{cases}\\2\sin x+2\sqrt3\cos x=0\\2\sin x=-2\sqrt3\cos x\\\sin x=-\sqrt3\cos x\\\tgx=\sqrt3\\x=\frac\pi3+\pi n,\;n\in\mathbb{Z}$
$x\in\left[\frac\pi2;\;2\pi\right]\Rightarrow\frac\pi2\leq\frac\pi 3+\pi n \leq2\pi\\\frac\pi6\leq\pi n\leq\frac{5\pi}3\\\frac16\leq n\leq\frac53\\n\in\mathbb{Z}\Rightarrow n=1\\x=\frac\pi3+\pi=\frac{4\pi}3$

Sin2x+2√3cos^x-6sinx-6√3cosx=0,** промежутке [pi/2;2*pi].