Question

1) sin(4x-п/6)=0 2) sin^2x-2sinxcosx=3cos^2x

Answer 1

$sin(4x-\frac{\pi}{6})=0\\4x-\frac{\pi}{6}=\pi n\\4x=\frac{\pi}{6}+\pi n\\x=\frac{\pi}{24}+\frac{\pi n}{4}, \; n\in Z;\\\\\\ sin^2x-2sinxcosx=3cos^2x\\sin^2x-2sinxcosx-3cos^2x=0|:cos^2x\\tg^2x-2tgx-3=0\\tgx=u\\u^2-2u-3=0\\D:4+12=16\\u=\frac{2\pm 4}{2}\\\\u_1=3\\tgx=3\\x=arctg3+\pi n, \; n\in Z;\\\\u_2=-1\\tgx=-1\\x=-\frac{\pi}{4}+\pi n, \; n\in Z.$

На cosx разделили при условии, что cosx ≠ 0, т.е. 
$cosx \neq 0\\x \neq \frac{\pi}{2}+\pi n, \; n\in Z$