Question

The value(s) of $\theta$, which satisfy $3-2cos\theta -4sin\theta -cos2\theta +sin2\theta =0$ is/are

• A. $\theta =2n\pi ;n\in 1$
• B. $2n\pi +\frac{\pi }{2};n\in 1$
• C. $2n\pi -\frac{\pi }{2};n\in 1$
• D. $n\pi ;n\in 1$

Answer 1

Answer: (A), (B)

The correct options are
A $\theta =2n\pi ;n\in 1$
B $2n\pi +\frac{\pi }{2};n\in 1$

$3-2\cos \theta -4\sin \theta -\cos 2\theta +\sin 2\theta =0$

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

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

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

$\Rightarrow (\sin \theta -1)[\sin \theta -1+\cos \theta ]=0$

Case $1$: Either $\sin \theta =1$

$\Rightarrow \theta =2n\pi +\pi /2$ where $n\in l$

Case $2$: Or, $\sin \theta +\cos \theta =1$

$\Rightarrow \cos (\theta -\pi /4)=\cos \left(\pi /4\right)$

$\Rightarrow \theta -\pi /4=2n\pi \pm \pi /4$

$\Rightarrow \theta =2n\pi ,2n\pi +\pi /2$ where $n\in l$

Hence, $\theta =2n\pi ,2n\pi +\pi /2$.