Question

The number of ordered pairs $\left(x,y\right)$ of real numbers satisfying the system of equations $sinx=sin2y$ and $cosx=siny$ , where $0\le x,y\le \pi$ , is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$\sin x=\sin 2y$ and $\cos x=\sin y$
Squaring and adding, we get, $1=(2\sin y\cos y{)}^2+{\sin }^{2}y$
$1-{\sin }^{2}y=4{\sin }^{2}y{\cos }^{2}y$
${\cos }^{2}y\left(1-4{\sin }^{2}y\right)=0$
$\cos y=0$ or $\sin y=\pm \frac{1}{2}$
$y=\frac{\pi }{2}$ or $\frac{\pi }{6}$ or $\frac{5\pi }{6}$
$y=\frac{\pi }{2}\Rightarrow x=0$
$y=\frac{\pi }{6}\Rightarrow x=\frac{\pi }{3}$
$\Rightarrow$ two ordered pairs $\left(0,\frac{\pi }{2}\right)\&\left(\frac{\pi }{3},\frac{\pi }{6}\right)$