Question

2sin^2 х+sinхcosх-3cos^2 х=0

Answer 1

$2sin^2x+sinxcosx-3cos^2x=0\;|cos^2x\\2tg^2x+tgx-3=0\\tgx=u\\2u^2+u-3=0\\D:1+24=25\\u_1,_2=\frac{-1\pm5}{4}\\u_1=1\\u_2=-\frac{3}{2}\\\\tgx=1\\x=\frac{\pi}{4}+\pi n,\;n\in Z\\\\tgx=-\frac{3}{2}\\x=-arctg\frac{3}{2}+\pi n,\;n\in Z$