Question

If $\sin \theta ,1,\cos 2\theta$ are in G.P., then

• A. $\theta =2n\pi +(-1{)}^n\frac{\pi }{2},n\in Z$
• B. $\theta =n\pi -\frac{\pi }{2},n\in Z$
• C. $\theta =2n\pi -\frac{\pi }{2},n\in Z$
• D. $\theta =n\pi +(-1{)}^n\frac{\pi }{2},n\in Z$

Answer 1

The correct option is C $\theta =2n\pi -\frac{\pi }{2},n\in Z$

$\sin \theta ,1,\cos 2\theta$ are in G.P.
If $a,b,c$ are in G.P., then
$\frac{b}{a}=\frac{c}{b}$
Now,
$\frac{1}{\sin \theta }=\frac{\cos 2\theta }{1}\Rightarrow \sin \theta \cos 2\theta =1\Rightarrow \sin \theta (1-2{\sin }^2\theta )=1\Rightarrow 2{\sin }^3\theta -\sin \theta +1=0$

By observation, $\sin \theta =-1$ is one root,
$\Rightarrow (\sin \theta +1)(2{\sin }^2\theta -2\sin \theta +1)=0\Rightarrow (\sin \theta +1)({\sin }^2\theta -\sin \theta +\frac{1}{2})=0\Rightarrow (\sin \theta +1)[{(\sin \theta -\frac{1}{2})}^2+\frac{1}{4}]=0\Rightarrow \sin \theta =-1\Rightarrow \theta =(4n-1)\frac{\pi }{2},n\in Z$