Решите два уравнения: 1) √3sin²x - 4sinxcosx + √3cos²x = 0 2) sinxcosx = -0.25

Решите задачу:

$1)\quad \sqrt3sin^2x-4sinx\cdot cosx+\sqrt3cos^2x=0\; |:cos^2x\ne 0\\\\\sqrt3tg^2x-4tgx+\sqrt3=0\\\\t=tgx\; ,\; \; \sqrt3t^2-4t+\sqrt3=0\\\\D/4=4-\sqrt3\cdot \sqrt3=4-3=1\\\\t_1=\frac{2-1}{\sqrt3}= \frac{1}{\sqrt3} =\frac{\sqrt3}{3}\; ;\; \; t_2= \frac{2+1}{\sqrt3} = \frac{3}{\sqrt3} =\sqrt3\\\\a)\quad tgx=\frac{\sqrt3}{3}\; ,\; \; x=\frac{\pi}{6}+\pi n,\; n\in Z\\\\b)\quad tgx=\sqrt3\; ,\; \; x=\frac{\pi}{3}+\pi k,\; k\in Z$

$2)\quad sinx \cdot c osx=-0,25\\\\\frac{1}{2}sin2x=-\frac{1}{4}\\\\sin2x=-\frac{1}{2}\\\\2x=(-1)^{n}arcsin(-\frac{1}{2})+\pi n\\\\2x=(-1)^{n}(-\frac{\pi}{6})+\pi n=(-1)^{n+1}\frac{\pi}{6}+\pi n\\\\x=(-1)^{n+1}\frac{\pi}{12}+\frac{\pi n}{2}\; ,\; n\in Z$