${\cos }^{-1}\left(\frac{3+5\cos x}{5+3\cos x}\right)=$

$\Rightarrow \cos \theta =\frac{3+5\cos x}{5+3\cos x}$  

$\Rightarrow \frac{1+\cos \theta }{1-\cos \theta }=\frac{5+3\cos x+3+5\cos x}{5+3\cos x-3-5\cos x}$

$\Rightarrow \frac{1+\cos \theta }{1-\cos \theta }=\frac{8\left(1+\cos x\right)}{2\left(1-\cos x\right)}$

$\Rightarrow {\cot }^2\frac{\theta }{2}=4{\cot }^2\frac{x}{2}$

$\Rightarrow {\tan }^2\frac{\theta }{2}=\frac{1}{4}{\tan }^2\frac{x}{2}$

$\Rightarrow \tan \frac{\theta }{2}=\frac{1}{2}\tan \frac{x}{2}$

$\Rightarrow \theta =2{\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$