Question

6sin^2x+4sin(pi-x)cos(2pi-x)=1

Answer 1

$6sin^2x+4sinx*cosx=1 \\6sin^2x+4sinxcosx=sin^2x+cos^2x \\5sin^2x+4sinxcosx-cos^2x=0 \\5*\frac{sin^2x}{cos^2x}+4*\frac{sinx}{cosx}-1=0 \\5tg^2x+4tgx-1=0 \\tgx=y \\5y^2+4y-1=0 \\D=16+20=36=6^2 \\y_1=\frac{-4+6}{10}=0,2 \\y_2=\frac{-4-6}{10}=-1 \\tgx=-1 \\x_1=-\frac{\pi}{4}+\pi n,\ n \in Z \\tgx=0,2 \\x_2=arctg(0,2)+\pi n,\ n \in Z$