Question

The solution of the system of equations $x+y=\frac{\pi }{2},\sin x+\sin y=\sqrt{2}$ is $\{x,y\}$, then

• A. $x=2k\pi +\frac{\pi }{4},y=\frac{\pi }{4}-2k\pi$
• B. $x=2k\pi -\frac{\pi }{2},y=\frac{\pi }{2}+x$
• C. $x=2k\pi +\frac{\pi }{3},y=\frac{\pi }{3}-2k\pi$
• D. $x=k\pi +\frac{\pi }{4},y=\frac{\pi }{4}-k\pi \left(\forall kϵI\right)$

Answer 1

The correct option is A $x=2k\pi +\frac{\pi }{4},y=\frac{\pi }{4}-2k\pi$

Given $x+y=\frac{\pi }{2}$ ....(1)

Also given $\sin x+\sin y=\sqrt{2}$

$\Rightarrow 2\sin \frac{x+y}{2}\cos \frac{x-y}{2}=\sqrt{2}$

$\Rightarrow \cos \frac{x-y}{2}=1\left(using\left(i\right)\right)$

$\Rightarrow x-y=2.2k\pi =4k\pi ,k\in z$

$\Rightarrow x-y=4\pi k$ ....(2)

Solving (1) and (2), we get
$\Rightarrow x=2k\pi +\frac{\pi }{4},kϵz$

$\Rightarrow y=\frac{\pi }{4}-2k\pi ,kϵz$