We have, sin 2x + cos x = 0 ⇒ 2 sinx cosx + cosx = 0 ⇒ cosx (2 sinx + 1) = 0 ⇒ cosx = 0 or, 2 sinx + 1= 0 Now if cosx = 0⇒ x = (2n + 1) $\frac{\pi}{2}$ , n ∊ Z And if 2 sinx + 1 = 0 ⇒ 2 sinx = – 1 ⇒ sinx = $-\frac{1}{2}$ A value of x satisfying sinx = $\frac{1}{2}$ is $\frac{\pi}{6}$ We have, sinx = $-\frac{1}{2}$ Thus, sin x = sin $\left(\pi+\frac{\pi}{6}\right)$ ⇒ sin x = sin $\frac{7\pi}{6}$ ⇒ x = nπ + $(-1)^n \frac{7\pi}{6}$ , n ∊ Z Hence x = (2n + 1) $\frac{\pi}{2}$ or nπ + $(-1)^n \frac{7\pi}{6}$ , n ∊ Z