If the sum of solutions of the system of equations 2 ^ 2 θ - 2θ = 0...

Given,

$2\sin^{2}\theta - \cos 2\theta = 0$

$⇒ 2\sin^{2}\theta - \left(1 - 2\sin^{2}\theta\right) = 0$

$⇒ \sin^{2}\theta = {\left(\dfrac{1}{2}\right)}^{2}$

$⇒ \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

Also given, $2\cos^{2}\theta + 3sin\theta = 0$

$⇒ 2\sin^{2}\theta - 3sin\theta - 2 = 0$

$⇒ sin\theta = - \dfrac{1}{2}$

$⇒ \theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

So, the common solution is

$\theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

Sum $= \dfrac{7\pi + 11\pi}{6} = 3\pi = k\pi$

$⇒ K = 3$