11Solving 2cos²x−cosx−1=0

Solve $\displaystyle 2\cos^2 x - \cos x - 1 = 0$ for $\displaystyle 0^\circ \le x < 360^\circ$.

The solutions are x = 0°, 120°, and 240°. Treating cos x as u gives (2u+1)(u−1)=0, so cos x = 1 (x=0°) or cos x = −1/2 (x=120°, 240°, since cosine is negative in quadrants II and III with reference angle 60°).