Question

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

Answer 1

${\tan }^2\theta =\frac{1-\cos 2\theta }{1+\cos 2\theta }=\frac{2}{4}=\frac{1}{2}$
$\therefore {\tan }^8\theta =\frac{1}{16}$. Put for ${\tan }^8\theta$
$\therefore 2{\cos }^2\alpha -3\cos \alpha -2=0$
$\therefore \cos \alpha =-\frac{1}{2},2$ (rejected)
$\therefore \cos \alpha =\cos (\pi -\frac{\pi }{3})=\cos \frac{2\pi }{3}$
$\therefore \alpha =2n\pi \pm \frac{2\pi }{3}$.