If $6\cos 2\theta +2{\cos }^2\left(\frac{\theta }{2}\right)+2{\sin }^2\theta =0, -\pi <\theta <\pi ,$ then $\theta =$

Given, $\theta \in \left(-\mathrm{\pi },\mathrm{\pi }\right)$

$6\cos 2\theta +2{\cos }^2\frac{\theta }{2}+2{\sin }^2\theta =0$

$\Rightarrow 6\left(1-2{\sin }^2\theta \right)+2{\cos }^2\frac{\theta }{2}+2{\sin }^2\theta =0$

$\Rightarrow 6-10{\sin }^2\theta +2{\cos }^2\left(\frac{\theta }{2}\right)=0$

$\Rightarrow 6-10\left(1-{\cos }^2\theta \right)+\left(1+\cos \theta \right)=0$

$\Rightarrow 10{\cos }^2\theta +\cos \theta -3=0$

$\Rightarrow 10{\cos }^2\theta -5\cos \theta +6\cos \theta -3=0$

$\Rightarrow \left(5\cos \theta +3\right)\left(2\cos \theta -1\right)=0$

$\Rightarrow \cos \theta =\frac{-3}{5},\frac{1}{2}$

$\Rightarrow \theta =\frac{\mathrm{\pi }}{3},\frac{-\mathrm{\pi }}{3},\mathrm{\pi }-{\cos }^{-1}\left(\frac{3}{5}\right),-\mathrm{\pi }+{\cos }^{-1}\left(\frac{3}{5}\right) \left\{\because \cos (-x)=\cos x\right\}$