The number of values of $\theta$  in the range $0\le \theta \le 360°$  satisfying the equation ${\cos }^{4}2\theta +2{\sin }^{2}2\theta =17{\left(\sin \theta +\cos \theta \right)}^8$  is

Given equation is ${(1-{\sin }^{2}2\theta )}^2+2{\sin }^{2}2\theta =17{(1+\sin 2\theta )}^4$ 

Put $x=\sin 2\theta .$  Therefore ${(1-x^2)}^2+2x^2=17{(1+x)}^4$

$x^4+1=17{\left(1+x\right)}^4=17{\left(1+2x+x^2\right)}^2$ ]

Clearly $x=\sin 2\theta =0$  is not a solution of the given equation. Therefore dividing both sides of Eq.  by $x^2$  we get $x^2+\frac{1}{x^2}=17{\left(x+\frac{1}{x}+2\right)}^2$

Putting $x+\frac{1}{x}=y,$  we have $y^2-2=17{\left(y+2\right)}^2$

$16y^2+68y+70=0,$

 $\Rightarrow 8y^2+34y+35=0$

   $\Rightarrow (4y+7)(2y+5)=0$

From this we get    $y=\frac{-7}{4}$  or $\frac{-5}{2}$

Case 1: If $y=-7/4,$  then $4x^2+7x+4=0$  has no real roots.

Case 2: If $y=-5/2,$  then

$$\begin{gathered} 2x^2+5x+2=0 \\ \Rightarrow x=\frac{-1}{2}, -2 \end{gathered}$$

Now $\sin 2\theta =x=-2$  is not possible . Therefore $\sin 2\theta =-1/2$

We get total 4 solutions