${\cos }^{4}2\theta +2{\sin }^{2}2\theta =17(\cos \theta +\sin \theta {)}^8$ if $\theta =$

${\left(1-{\sin }^{2}2\theta \right)}^2+2{\sin }^{2}2\theta =17(1+\sin 2\theta {)}^4$
$\Rightarrow 1+x^4=17(1+x{)}^4,$ where $x=\sin 2\theta$

 or  $x^2+\frac{1}{x^2}=17{\left(x+\frac{1}{x}+2\right)}^2$

If  $t=x+\frac{1}{x},$ we get $t^2-2=17(t+2{)}^2$
$\Rightarrow 8t^2+34t+35=0\therefore x+\frac{1}{x}=t=-\frac{5}{2},-\frac{7}{4}$
But $x+\frac{1}{x}$ cannot lie between -2 and 2,

$\begin{array}{r}x+\frac{1}{x}=-\frac{5}{2}\Rightarrow x=\sin 2\theta =-\frac{1}{2}\\ \therefore \theta ={105}^{\circ },{165}^{\circ },{285}^{\circ },{345}^{\circ }\end{array}$