Question

2cos в квадрате x+9 cosx+4=0

Answer 1

$2Cos^{2}x+9Cosx+4=0\\\\Cosx=m,-1\leq m\leq1\\\\2m^{2}+9m+4=0\\\\D=9^{2}-4*2*4=81-32=49=7^{2}\\\\m_{1}=\frac{-9+7}{4}=-\frac{1}{2}\\\\m_{2}=\frac{-9-7}{4}=-4$

m₂ = - 4 - не подходит

$Cosx=-\frac{1}{2}\\\\x=\pm arcCos(-\frac{1}{2})+2\pi n,n\in Z\\\\x=\pm \frac{2\pi }{3}+2\pi n,n\in Z$