Question

The number of ordered pairs $\left(x , y\right)$ of real numbers satisfying the system of equations $sin x=sin ⁡ 2 y$ and $cos x=sin ⁡ y$ , where $0\leq x,y\leq \pi $ , is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$\sin x=\sin 2 y$ 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 $\left(0, \frac{\pi}{2}\right) \&\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$