Question 384531
If you let z = cos x, then the equation becomes a simple quadratic:

{{{z^2 - z - 2 = 0}}} (I hope the cos2x means {{{cos^2 (x)}}}, otherwise we'd need the power reduction formulas).

From the quadratic, we can factor and obtain {{{(z - 2)(z + 1) = 0}}} --> z = 2 or z = -1. However, z is only defined on [-1, 1] since it is the range of cosine, so z = -1, and x = -pi (plus any multiple of 2pi since they denote the same angle on a unit circle).