Question

2cos^2x+корень из 3sin2x+1=0 [-п/2; 2п]

ПОМОГИИИИИТЕ ПОЖААААЛУЙСТА

Answer 1

$2\cos^2x+\sqrt3\sin2x+1=0\\2\cos^2x+2\sqrt3\sin x\cos x+\sin^2x+\cos^2x=0\\3\cos^2x+2\sqrt3\sin x\cos x+\sin^2x=0\\(\sqrt3\cos x)^2+2\sqrt3\cos x\sin x+(\sin x)^2=0\\(\sqrt3\cos x+\sin x)^2=0\\\sqrt3\cos x+\sin x=0\\\sin x=-\sqrt3\cos x\\\frac{\sin x}{\cos x}=-\sqrt3\\tgx=-\sqrt3\\x=\frac{2\pi}3+\pi k,\;k\in\mathbb{Z}$

$x\in[-\frac\pi2;\;2\pi]:\\\\-\frac\pi2\leq\frac{2\pi}3+\pi k\leq2\pi\\\\-\frac\pi2-\frac{2\pi}3\leq\pi k\leq2\pi-\frac{2\pi}3\\\\-\frac{5\pi}6\leq\pi k\leq\frac{4\pi}3\\\\-\frac56\leq k\leq\frac43\\\\k\in\mathbb{Z}\Rightarrow k=0,\;1$

Корни, принадлежащие заданному отрезку:

$\boxed{x_1=\frac{2\pi}3;\;x_2=\frac{2\pi}3+\pi=\frac{5\pi}3}$