Question

2cos2x-3cos²x-2sinx=0​

Answer 1

$\displaystyle\bf\\2Cos2x-3Cos^{2} x-2Sinx=0\\\\2\cdot(1-2Sin^{2} x)-3\cdot(1-Sin^{2} x)-2Sinx=0\\\\2-4Sin^{2} x-3+3Sin^{2} x-2Sinx=0\\\\-Sin^{2} x-2Sinx-1=0\\\\Sin^{2} x+2Sinx+1=0\\\\Sinx=m \ , \ -1\leq m\leq 1\\\\m^{2} +2m+1=0\\\\(m+1)^{2} =0\\\\m+1=0 \ \ \Rightarrow \ \ m=-1\\\\Sinx=-1\\\\\boxed{x=-\frac{\pi }{2} +2\pi n \ , \ n\in Z}$