Salvatore Boone

2021-12-24

How do you solve $\mathrm{sin}2x=\mathrm{sin}x$?

Philip Williams

Expert

$\mathrm{sin}2x=\mathrm{sin}x$
$2\mathrm{sin}x\mathrm{cos}x=\mathrm{sin}x$
$2\mathrm{sin}x\mathrm{cos}x-\mathrm{sin}x=0$
$\mathrm{sin}x\left(2\mathrm{cos}x-1\right)=0$
Solution A: $\mathrm{sin}x=0⇒x=k\pi ,k\in \mathbb{Z}$
Solution B:
$\therefore x=k\pi$ or $x=\frac{\pi }{3}\left(6k±1\right),k\in \mathbb{Z}$

Paineow

Expert

Factor $\mathrm{sin}\left(x\right)$ out of $2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-\mathrm{sin}\left(x\right)$
$\mathrm{sin}\left(x\right)\left(2\mathrm{cos}\left(x\right)-1\right)=0$
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
$\mathrm{sin}\left(x\right)=0$
$2\mathrm{cos}\left(x\right)-1=0$
Set the first factor equal to 0 and solve.
$x=2\pi n,\pi +2\pi n$ for any integer n
Set the next factor equal to 0 and solve.
$x=\frac{\pi }{3}+2\pi n$, for any integer n
The final solution is all the values that make $\mathrm{sin}\left(x\right)\left(2\mathrm{cos}\left(x\right)-1\right)=0$ true.
$x=2\pi n,\pi +2\pi n,\frac{\pi }{3}+2\pi n,\frac{5\pi }{3}+2\pi n$, for any integer n
The finall solution is all the values that make $\mathrm{sin}\left(x\right)\left(2\mathrm{cos}\left(x\right)-1\right)=0$ true.
$x=2\pi n,\pi +2\pi n,\frac{\pi }{3}+2\pi n,\frac{5\pi }{3}+2\pi n$, for any integer n
Consolidate $2\pi n$ and $\pi +2\pi n$ to $\pi n$
$x=\pi n,\frac{\pi }{3}+2\pi n,\frac{5\pi }{3}+2\pi n$ for any integer n

user_27qwe

Expert