Solve using factoring

Answer:
[tex]{ \tt{4 { \cos }^{2} x - 1 = 0}} \\ { \tt{ {(2 \cos x) }^{2} - {1}^{2} = 0 }} \\ { \tt{(2 \cos x - 1)(2 \cos x + 1) = 0}} [/tex]
» Either 2cos x -1 or 2cos x +1 is zero
[tex]{ \tt{2 \cos x = \pm 1 }} \\ { \tt{ \cos x = \pm \frac{1}{2} }} \\ \\ { \tt{x = 60 \degree \: and \: 120 \degree}}[/tex]