Помогите решить уравнения а)sinx-2sinxcosx+4cosx-2=0 b)3sin^2x=2sinxcosx+cos^2x...

Решите задачу:

$sinx-2sinxcosx+4cosx-2=0\\sinx(1-2cosx)-2(1-cosx)=0\\(sinx-2)(1-2cosx)=0\\\\sinx-2=0\\sinx \neq 2\\sinx\in [-1;1]; \\\\1-2cosx=0\\cosx=\frac{1}{2}\\x=\pm \frac{\pi}{3}+2\pi n, \; n \in Z;$

$3sin^2x=2sinxcosx+cos^2x\\3sin^2x-2sinxcosx-cos^2x=0|:cos^2x\\3tg^2x-2tgx-1=0\\tgx=u\\3u^2-2u-1=0\\D:4+12=16\\u=\frac{2\pm 4}{6}\\\\u_1=1\\tgx=1\\x=\frac{\pi}{4}+\pi n. \; n\in Z;\\\\u_2=-\frac{1}{3}\\tgx=-\frac{1}{3}\\x=-arctg\frac{1}{3}+\pi n, \; n\in Z\\\\ cosx \neq 0\\x \neq \frac{\pi}{2}+\pi k, \; k\in Z$

$5sin^2x-2sinxcosx+cos^2x=4\\5sin^2x-2sinxcosx+cos^2x-4sin^2x-4cos^2x=0\\sin^2x-2sinxcosx-3cos^2x=0| cos^2x\\tg^2x-2tgx-3=0\\tgx=u\\u^2-2u-3=0\\D:4+12=16\\u=\frac{2\pm 4}{2}\\\\u_1=3\\tgx=3\\x=arctg3+\pi n, \; n\in Z;\\\\u_2=-1\\tgx=-1\\x=-\frac{\pi}{4}+\pi n, \; n\in Z;\\\\cosx \neq 0\\x \neq \frac{\pi}{2}+\pi k, \; k\in Z.$