A Quadratic in Sine

Solve $2{\cos\left(x\right)}^{2}-3\cos\left(x\right)+1 = 0$. With u for cos x it is $2{u}^{2}-3u+1 = 0$, which factors as $\left(2u-1\right)\left(u-1\right)$: multiply back to check, $2{u}^{2}-2u-u+1$.

Put the cosine back and split: 2cos x − 1 = 0 gives cos x = 1/2, so 60° or 300°, and cos x − 1 = 0 gives cos x = 1, so 0° only, since 360° is outside the domain. Three answers.