Question

If $32ta{n}^8\theta =2co{s}^2\alpha -3cos\alpha and3cos2\theta =1$, then find the general value of $\alpha$

• A. $\alpha =2n\pi \pm \frac{2\pi }{3}$
• B. $\alpha =2n\pi +\frac{2\pi }{3}$
• C. $\alpha =2n\pi \pm \frac{\pi }{3}$
• D. $\alpha =2n\pi \pm \frac{2\pi }{6}$

Answer 1

The correct option is A $\alpha =2n\pi \pm \frac{2\pi }{3}$

Given $3\cos 2\theta =1\Rightarrow \cos 2\theta =\frac{1}{3}$
${\tan }^2\theta =\frac{1-\cos 2\theta }{1+\cos 2\theta }=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{1}{2}$
Now $32{\tan }^8\theta =2{\cos }^2\alpha -3\cos \alpha$
$\Rightarrow 2{\cos }^2\alpha -3\cos \alpha =32{\left(\frac{1}{2}\right)}^4=2$
$\Rightarrow 2{\cos }^2\alpha -3\cos \alpha -2=0$

$\Rightarrow (\cos \alpha -2)(2\cos \alpha +1)=0$
$\Rightarrow \cos \alpha =-\frac{1}{2}[\because \cos \alpha \ne 2]$
$\Rightarrow \alpha =2n\pi \pm \frac{2\pi }{3}$