Question

2 sin^2 x +3 cos x - 3 = 0

Answer 1

$\displaystyle 2sin^2x+3cosx-3=0\\2(1-cos^2x)+3cosx-3=0\\2-2cos^2x+3cosx-3=0\\2cos^2x-3cosx+1=0\\(cosx)_{1,2}=\frac{3\pm\sqrt{9-8}}{4}=\frac{3\pm1}{4}\\cosx_1=1\ \ \ \ \ \ \ \ \ \ \ \ cosx_2=\frac{1}{2}\\x_1=2\pi n;\ \ \ \ \ \ \ \ \ \ \ x_2=\pm\frac{\pi}{3}+2\pi n;n\in Z$