Question

If $x+y=\frac{2\pi }{3}$ and $\frac{\sin x}{\sin y}=2$ , then

• A. the number of values of $x\in [0,4\pi ]$ are $4$
• B. the number of values of $x\in [0,4\pi ]$ are $2$
• C. the number of values of $y\in [0,4\pi ]$ are $4$
• D. the number of values of $y\in [0,4\pi ]$ are $8$

Answer 1

Answer: (A), (C)

The correct options are
A the number of values of $x\in [0,4\pi ]$ are $4$
C the number of values of $y\in [0,4\pi ]$ are $4$

$x+y=\frac{2\pi }{3}\Rightarrow y=\frac{2\pi }{3}-x$
$\therefore \sin x=2\sin (\frac{2\pi }{3}-x)$
$\Rightarrow \sin x=2[\left(\frac{\sqrt{3}}{2}\right)\cos x+\frac{1}{2}\sin x]\Rightarrow \sin x=\sqrt{3}\cos x+\sin x\Rightarrow \cos x=0\Rightarrow x=\frac{(2n+1)\pi }{2},n\in Z$
$\Rightarrow y=\frac{2\pi }{3}-n\pi -\frac{\pi }{2}$
$\Rightarrow y=\frac{\pi }{6}-n\pi ,n\in Z$

Hence, for $x\in [0,4\pi ],$
$x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}$

and for $y\in [0,4\pi ],$
$y=\frac{\pi }{6},\frac{7\pi }{6},\frac{13\pi }{6},\frac{19\pi }{6}$