Question

4cos²x-2sin²x-5cosx-4=0

Answer 1

$4Cos^{2}x-2Sin^{2}x-5Cosx-4=0\\\\4Cos^{2}x-2(1-Cos^{2}x)-5Cosx-4=0\\\\4Cos^{2}x-2+2Cos^{2}x-5Cosx-4=0\\\\6Cos^{2}x-5Cosx-6=0\\\\Cosx=m,-1 \leq m \leq 1 \\\\6m^{2}-5m-6=0\\\\D=(-5)^{2}-4*6*(-6)=25+144=169=13^{2}\\\\m_{1}=\frac{5-13}{12}=-\frac{8}{12}=-\frac{2}{3} \\\\m_{2}=\frac{5+13}{12}=\frac{18}{12}=1,5>1-neyd\\\\Cosx=-\frac{2}{3}\\\\x=\pm arc Cos(-\frac{2}{3})+2\pi n,n\in Z$