sec x = 2 ⇒ cos x = $\frac{1}{2}$ A value of x satisfying cosx = $\frac{}{2}$ is $\frac{\pi}{3}$ Principal Solution: We know that, cos $\frac{\pi}{3}$ = $\frac{1}{2}$ and cos $\left(2\pi - \frac{\pi}{3}\right)$ = cos $\frac{\pi}{3}$ = $\frac{1}{2}$ ⇒ Thus, cos $\frac{\pi}{3}$ = cos $\frac{5\pi}{3}$ = $\frac{1}{2}$ Therefore principal solutions are $\frac{\pi}{3}$ , $\frac{6\pi}{3}$ General Solution : We have, cos x = $\frac{1}{2}$ ⇒ cos x = cos $\frac{\pi}{3}$ ⇒ x = 2nπ ± $\frac{\pi}{3}$ , n ∊ Z Since If cos θ = cos α , then θ = 2nπ ± α , n ∊ Z