Question

Solve using factoring

Answer 1

`{ \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}}`

» Either 2cos x -1 or 2cos x +1 is zero

`{ \tt{2 \cos x = \pm 1 }} \\ { \tt{ \cos x = \pm (1)/(2) }} \\ \\ { \tt{x = 60 \degree \: and \: 120 \degree}}`