We have, cos 3x + cos x – cos 2x = 0 ⇒ 2 cos $\left(\frac{3x+x}{2}\right)$ cos $\left(\frac{3x-x}{2}\right)$ - cos 2x = 0 ⇒ 2cos 2x cosx – cos 2x = 0 ⇒ cos 2x (2 cosx – 1) = 0 Either cos 2x = 0 or, 2 cosx – 1 = 0 Now, if cos 2x = 0 ⇒ 2x = (2x + 1) $\frac{\pi}{2}$ ,n ∊ Z ⇒ x = (2n + 1) $\frac{\pi}{4}$ , n ∊ Z [Since cos x = 0 , then x = (2n + 1) $\frac{\pi}{2}$ , n ∊ Z] If, 2 cosx – 1 = 0 ⇒ 2 cosx = 1 ⇒ cosx = 1/2 ⇒ cos x = cos $\frac{\pi}{3}$ ⇒ x = 2nπ ± $\frac{\pi}{3}$ , n ∊ Z Since if cos x = cos α , then x = 2nπ ± $\frac{\pi}{3}$ , n ∊ Z Here x = (2n + 1) $\frac{\pi}{4}$ or 2nπ ± $\frac{\pi}{3}$ , n ∊ Z