Question

The general solution(s) of $\theta$ which satisfy $3-2\cos \theta –4\sin \theta -\cos 2\theta +\sin 2\theta =0$ is/are (where $n\in Z$)

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

Answer 1

Answer: (A), (B)

The correct options are
A $2n\pi$
B $2n\pi +\frac{\pi }{2}$

$3-2\cos \theta -4\sin \theta -\cos 2\theta +\sin 2\theta =0$ $\Rightarrow 3-2\cos \theta -4\sin \theta -1+2{\sin }^2\theta +2\sin \theta \cos \theta =0$
$\Rightarrow 2{\sin }^2\theta -2\cos \theta -4\sin \theta +2\sin \theta \cos \theta +2=0$
$\Rightarrow ({\sin }^2\theta -2\sin \theta +1)+\cos \theta (\sin \theta -1)=0$
$\Rightarrow (\sin \theta -1)[\sin \theta -1+\cos \theta ]=0$

Case $1:$ Either $\sin \theta =1$
$\Rightarrow \theta =2n\pi +\frac{\pi }{2}\text{where}n\in Z$

Case $2:$ $\sin \theta +\cos \theta =1$
$\Rightarrow \cos (\theta -\frac{\pi }{4})=\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}\Rightarrow \theta -\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4}\Rightarrow \theta =2n\pi ,2n\pi +\frac{\pi }{2}$
Hence, $\theta =2n\pi ,2n\pi +\frac{\pi }{2},n\in Z$