$2\cos \mathrm{\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-16{\cos }^3{\mathrm{\theta sin}}^2\mathrm{\theta }=$

$\begin{array}{l}=2\cos \mathrm{\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-2\left(4{\cos }^3\mathrm{\theta }\right)\left(2{\sin }^2\mathrm{\theta }\right)\\ =2\cos \mathrm{\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-2(\cos 3\mathrm{\theta }+3\cos \mathrm{\theta })(1-\cos 2\mathrm{\theta })\\ =2\cos \mathrm{\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-2\cos 3\mathrm{\theta }+2\cos 3\mathrm{\theta cos}2\mathrm{\theta }-6\cos \mathrm{\theta }+6\cos \mathrm{\theta cos}2\mathrm{\theta }\\ =-4\cos \mathrm{\theta }-3\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }+\cos 5\mathrm{\theta }+\cos \mathrm{\theta }+3(\cos 3\mathrm{\theta }+\cos \mathrm{\theta })=0\end{array}$