If $32{\tan }^8\theta =2{\cos }^2\alpha -3\cos \alpha$  and $3\cos 2\theta =1$ , then the general value of $\alpha$ is

$$\begin{gathered} 3\cos 2\theta =1 \\ \Rightarrow {\tan }^2\theta =\frac{1-\cos 2\theta }{1+\cos 2\theta }=\frac{1}{2} \end{gathered}$$ 

$2{\cos }^2\alpha -3\cos \alpha =32\left(\frac{1}{16}\right)\Rightarrow 2{\cos }^2\alpha -3\cos \alpha -2=0$

$\left(2\cos \alpha +1\right)\left(\cos \alpha -2\right)=0\Rightarrow \cos \alpha =-\frac{1}{2}=\cos \frac{2\pi }{3}$

General solution is: $\alpha =2n\pi \pm \frac{2\pi }{3};n\in z$