Question

The value(s) of $0$, which satisfy $3-2\cos \theta -4\sin \theta -\cos 2\theta +\sin 2\theta =0$ 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: (B)

$2n\pi +\frac{\pi }{2};n\in 1$

Mistake : ans is A,B

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

$let\sin \theta =s,\cos \theta =c⟹\sin 2\theta =2cs,\cos 2\theta =2c^2-1$

$3-2c-4s-2c^2+1+2cs=0⟹sc-2s=c+c^2-2$

squaring on both sides gives

$s^2{c}^2+4s^2-4s^{2}c=c^4+c^2+4-4c-4c^2+2c^3$

using $s^2=1-c^2⟹(1-c^2)c^2+4(1-c^2)-4(1-c^2)c=c^{}+c^{}+4-4c-4c^2+2c^3$

on simplifying gives

$c^3(c-1)=0⟹c=0,1⟹s=0,-1,1$

but $s=-1$ is not satisfting therefore values satisfying are $c=1,s=1⟹\theta =2n\pi ,2n\pi +\frac{\pi }{2};n\in I$