Question

Помогите решить... ​

Answer 1

Ответ:

$2sin^{2} x+7cosx+2=0\\\\4+7cosx-2cosx^{2} =0\\\\-(cosx-4)(2cosx+1)=0\\\\(cosx-4)(2cosx+1)=0\\\\cosx-4=0; 2cosx+1=0\\\\cosx=4; 2cosx=-1\\\\cosx=-\frac{1}{2} \\\\x= 2\pi n_{1} +\frac{2\pi }{3}$