If $\sqrt[3]{x+iy}={\cos }^2\theta +i{\sin }^2\theta ,$then minium value of $\sin 4\theta$ is----

$\left(x+iy\right)={\left({\cos }^2\theta +i{\sin }^2\theta \right)}^3$

$\Rightarrow x={\cos }^6\theta -3{\cos }^2\theta {\sin }^4\theta$  and $y=3{\cos }^4\theta {\sin }^2\theta -{\sin }^6\theta$

$\Rightarrow \frac{x}{{\cos }^2\theta }={\cos }^4\theta -3{\sin }^4\theta ,{\sin }^4\theta ,\frac{y}{{\sin }^2\theta }=3{\cos }^4\theta -{\sin }^4\theta$

Adding   $\Rightarrow x{\sec }^2\theta +y\cos e{c}^2\theta =4\left({\cos }^4\theta -{\sin }^4\theta \right)--------\left(1\right)$

Now  $AM\ge GM\Rightarrow \frac{x{\sec }^2\theta +y\cos e{c}^2\theta }{2}\ge \sqrt{x{\sec }^2\theta y\cos e{c}^2\theta }$

$\Rightarrow \frac{4\left({\cos }^4\theta -{\sin }^4\theta \right)}{2}\ge \frac{\sqrt{xy}}{\sin \theta \cos \theta }\Rightarrow 2\cos 2\theta \ge \frac{2\sqrt{xy}}{\sin 2\theta }$

$\Rightarrow \sin 4\theta \ge 2\sqrt{xy}$