Question

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

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

Answer 1

Answer: (B)

$2n\pi \pm \frac{2\pi }{3}$

Given,

$32{\tan }^8\theta =2{\cos }^2\alpha -3\cos \alpha$...............(i)

$3\cos 2\theta =1$

This gives $\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{1}{3}$

So, ${\tan }^2\theta =\frac{1}{2}$

Given equation (i) becomes

$32({\tan }^2\alpha {)}^4=2{\cos }^2\alpha -3\cos \alpha$

$\Rightarrow 32\times \frac{1}{16}=2{\cos }^2\alpha -3\cos \alpha$

$\Rightarrow 2=2{\cos }^2\alpha -3\cos \alpha$

$\Rightarrow 2{\cos }^2\alpha -3\cos \alpha -2=0$

$\Rightarrow 2{\cos }^2\alpha -4\cos \alpha +\cos \alpha -2=0$

$\Rightarrow 2\cos \alpha (\cos \alpha -2)+1(\cos \alpha -2)=0$
$\Rightarrow (2\cos \alpha +1)(\cos \alpha -2)=0$

So, $\cos \alpha =-\frac{1}{2}$ or $\cos \alpha =2$

$\because \cos \alpha \ne 2$

$\therefore \cos \alpha =-\frac{1}{2}\Rightarrow \alpha =\frac{2\pi }{3}$
So. general solution is $\alpha =2n\pi \pm \frac{2\pi }{3}$

Hence, the correct answer is B