Question

2cos в квадрате x+9 cosx+4=0

Answer 1

$2Cos^{2}x+9Cosx+4=0\\Cosx=m,-1leq mleq1\\2m^{2}+9m+4=0\\D=9^{2}-4*2*4=81-32=49=7^{2}\\m_{1}=frac{-9+7}{4}=-frac{1}{2}\\m_{2}=frac{-9-7}{4}=-4$

m₂ = - 4 - не подходит

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