$If 5\cos 2\theta +2{\cos }^2\frac{\theta }{2}+1, -\pi <\theta <\pi then \theta =$

$$\begin{gathered} 5\cos 2\mathrm{\theta }+2{\cos }^2\frac{\mathrm{\theta }}{2}+1=0 \\ 5\left(2{\cos }^2\mathrm{\theta }-1\right)+1+\mathrm{cos\theta }+1=0 \\ 10{\cos }^2\mathrm{\theta }-5+\mathrm{cos\theta }+2=0 \\ 10{\cos }^2\mathrm{\theta }+\mathrm{cos\theta }-3=0 \\ (5\mathrm{cos\theta }+3) (2\mathrm{cos\theta }-1)=0 \\ \mathrm{cos\theta }=\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right. \mathrm{cos\theta }=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{\theta }=\mathrm{\pi }-{\cos }^{-1}\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right), \pm \frac{\mathrm{\pi }}{3} \end{gathered}$$