Question

If $\alpha <\beta ,0\le \alpha ,\beta \le 2\pi$ and $\alpha ,\beta$ satisfy the equation $\sin 2x+3\sin x=1+3\cos x,$ then which of the following is/are true?

• A. $\alpha +\beta =\frac{\pi }{2}$
• B. $\alpha +\beta =\frac{3\pi }{2}$
• C. $\cos \alpha +\cos \beta =0$
• D. $\sin \alpha +\sin \beta =0$

Answer 1

Answer: (B), (C), (D)

$\sin \alpha +\sin \beta =0$

Given equation,
$\sin 2x+3\sin x=1+3\cos x,$
$\Rightarrow \sin 2x+3(\sin x-\cos x)-1=0\dots \left(1\right)$

Let $\sin x-\cos x=t$
$\Rightarrow \sin 2x=1-t^2$

From $\left(1\right),$
$1-t^2+3t-1=0\Rightarrow t^2-3t=0$
$\Rightarrow t=0\text{or}t=3$
$\Rightarrow \sin x-\cos x=0\text{or}\sin x-\cos x=3\text{(rejected)}$
$\Rightarrow \sin x=\cos x\Rightarrow \tan x=1=\tan \frac{\pi }{4}$
$\Rightarrow x=n\pi +\frac{\pi }{4},n\in Z$
$\Rightarrow \alpha =\frac{\pi }{4},\beta =\frac{5\pi }{4}(\because \alpha <\beta ,0\le \alpha ,\beta \le 2\pi )$

$\alpha +\beta =\frac{3\pi }{2}$

$\cos \alpha +\cos \beta =\cos \frac{\pi }{4}+\cos \frac{5\pi }{4}=0$

$\sin \alpha +\sin \beta =\sin \frac{\pi }{4}+\sin \frac{5\pi }{4}=0$