Question

6sin^2x+sinxcosx-cos^2x=0 решите уравнение

Answer 1

$6sin^2x+sinxcosx-cos^2x=0|:cos^2x\\6tg^2x+tgx-1=0\\tgx_{1,2}=\frac{-1^+_-5}{12}\\tgx_1=-\frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;tgx_2=\frac{1}{3}\\x_1=-arctg\frac{1}{2}+\pi n;n\in Z\ \ x_2=arctg\frac{1}{3}+\pi n;n\in Z$