Inverse Trigonometry doubt.

Question

Suppose $\sin y=\sin 2x$, then what will be the solution for $y$?

Will it be $y=2x$ or $y=n\pi-2x$ for some $n \in \mathbb{N}$?

Without additional information both are correct. – Git Gud Jun 13 '13 at 16:11 — Git Gud, Jun 13 '13 at 16:11

score 1 · Answer 1 · answered Jun 13 '13 at 16:12

1

HINT:

Use $\sin(\pi-2x)=\sin 2x$ and $\sin(2m\pi+z)=\sin z$ where $m$ is any integer

so, $y=2m\pi+2x$ and $y=2m\pi+\pi-2x=(2m+1)\pi-2x$

which can be merged as $y=n\pi+(-1)^n(2x)$ where $n$ is any integer

answered Jun 13 '13 at 16:12

lab bhattacharjee

Yes, thanks. I got that. – Adienl Jun 13 '13 at 16:16
2

@AdityaParson, thanks, nice to hear that. – lab bhattacharjee Jun 13 '13 at 16:19

1 Answers1