Question

4sin^2x-3sinxcosx=cos^2x

Answer 1

$4sin^2x-3sinx*cosx=cos^2x|:cos^2x \neq 0\\4tg^2x-3tgx=1\\4tg^2x-3tgx-1=0\\t=tgx\\4t^2-3t-1=0\\D=(-3)^2-4*4*(-1)=9+16=25=5^2\\t_1= \frac{3+5}{2*4}= \frac{8}{8}=1\\\\t_2= \frac{3-5}{2*4}= \frac{-2}{8}=-0,25\\\\tgx=1\\x_1= \pi /4+ \pi n, n\in Z\\\\tgx=-0,25\\x_2=-arctg0,25+ \pi n, n\in Z$