میدانیم ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ پس:
$2{{\sin }^{2}}x=4+5\operatorname{cosx}\Rightarrow 2\left( 1-{{\cos }^{2}}x \right) =4+5\operatorname{cosx}$
$\Rightarrow 2{{\cos }^{2}}x+5\operatorname{cosx}+2=0\Rightarrow \left( 2\operatorname{cosx}+1 \right)\left( \operatorname{cosx}+2 \right)=0$
$\Rightarrow \left\{ \begin{matrix} \operatorname{cosx}=-\frac{1}{2}=\cos \frac{2\pi }{3}\Rightarrow x=2k\pi \pm \frac{2\pi }{3}\left( k\in z \right) \\ \operatorname{cosx}=-2 \\ \end{matrix} \right.$