Question

3cos2x-7sin=4
Помогите решить

Answer 1

$cos{2x}=1-2sin^2x \\ 3*(1-2sin^2x)-7sinx-4=0 \\ 3-6sin^2x-7sinx-4=0 \\ 6sin^2x+7sinx+1=0 \\ sinx=t \\ 6t^2+7t+1=0 \\ t_1=- \frac{1}{6} \\ t_2= -1 \\ \left \{ {{sinx=-1} \atop {sinx=-\frac{1}{6} }} \right. \left \{ {{x_1= \frac{3 \pi }{2}+2\pi n } \atop {x_2=arcsin(-\frac{1}{6})+2\pi n }}\atop {x_3= \pi+arcsin(-\frac{1}{6})+2\pi n }} \right.$