Question

В)-2cos^2x-5sinx-1=0 г)sin^2x+4sinxcosx-5cos^2=0

Answer 1

В)
$-2cos^2x-5sinx-1=0 \\-2(1-sin^2x)-5sinx-1=0 \\sinx=y,\ y \in [-1;1] \\-2+2y^2-5y-1=0 \\2y^2-5y-3=0 \\D=25+24=49=7^2 \\y_1= \frac{5+7}{4} =3 \notin[-1;1] \\y_2= \frac{5-7}{4} =- \frac{1}{2} \in [-1;1] \\sinx=- \frac{1}{2} \\x_1= -\frac{\pi}{6} +2\pi n, \ n \in Z \\x_2=-\frac{5\pi}{6} +2\pi n, \ n \in Z$
г)
$sin^2x+4sinxcosx-5cos^2=0 \\ \frac{sin^2x}{cos^2x} -4* \frac{sinx}{cosx} -5=0 \\tg^2x-4tgx-5=0 \\tgx=y \\y^2-4y-5=0 \\D=16+20=36=6^2 \\y_1= \frac{4+6}{2} =5 \\y_2= \frac{4-6}{2} =-1 \\tgx=5 \\x_1=arctg(5)+\pi n,\ n \in Z \\tgx=-1 \\x_2=- \frac{\pi}{4} +\pi n,\ n \in Z$