Been having problem with a few questions from my review for solving trig equations
$\displaystyle
Cos2X(Sin2x-1)=0
$
and
$\displaystyle
2Cos^2x-cosx-1=0
$
$\displaystyle \cos(2x)[\sin(2x)-1]=0$
set each factor equal to 0 and solve ...
$\displaystyle \cos(2x) = 0$
$\displaystyle 2x = \frac{\pi}{2}$ , $\displaystyle 2x = \frac{3\pi}{2}$ , ...
$\displaystyle \sin(2x) = 1$
$\displaystyle 2x = \frac{\pi}{2}$ , $\displaystyle 2x = \frac{5\pi}{2}$ , ...
$\displaystyle 2\cos^2{x}-\cos{x}-1=0$
$\displaystyle (2\cos{x} + 1)(\cos{x} - 1) = 0$
same drill ... set each factor equal to 0 and solve.