Question

Решить уравнение 3sin²x-2sinxcosx =cos²x

Answer 1

$3sin^2x-2sinxcosx =cos^2x \\3sin^2x-2sinxcosx-cos^2x=0 \\ 3*\frac{sin^2x}{cos^2x} -2 \frac{sinx}{cosx} -1=0 \\3tg^2x-2tgx-1=0 \\tgx=y \\3y^2-2y-1=0 \\D=4+12=16=4^2 \\y_1= \frac{2+4}{6} =1 \\y_2= \frac{2-4}{6} =- \frac{1}{3} \\tgx=1 \\x_1= \frac{\pi}{4} +\pi n,\ n \in Z \\tgx=- \frac{1}{3} \\x_2=-arctg( \frac{1}{3} )+\pi n,\ n \in Z$