1. ## Trig equation

I'm kind of stuck on the final steps of this problem.

Determine all solution of the equation 5 sin 3x - 11= 3 sin 3x - 12 in the domain 0≤x≤2π

I've gotten to the point of sin 3x = -0.5, but I'm not sure of what to do next.

2. Originally Posted by DemonX01
I'm kind of stuck on the final steps of this problem.

Determine all solution of the equation 5 sin 3x - 11= 3 sin 3x - 12 in the domain 0≤x≤2π

I've gotten to the point of sin 3x = -0.5, but I'm not sure of what to do next.
$\displaystyle 0 \le x \le 2\pi$

$\displaystyle 0 \le 3x \le 6\pi$

let $\displaystyle u = 3x$ ...

$\displaystyle 0 \le u \le 6\pi$

$\displaystyle \sin{u} = -\frac{1}{2}$

unit circle values ...

$\displaystyle u = \frac{7\pi}{6} \, , \, \frac{11\pi}{6} \, , \, \frac{19\pi}{6} \, , \, \frac{23\pi}{6} \, , \frac{31\pi}{6} \, , \, \frac{35\pi}{6}$

$\displaystyle x = \frac{u}{3} = \frac{7\pi}{18} \, , \, \frac{11\pi}{18} \, , \, \frac{19\pi}{18} \, , \, \frac{23\pi}{18} \, , \frac{31\pi}{18} \, , \, \frac{35\pi}{18}$

3. Hello, DemonX01!

A variation of skeeter's solution . . .

Determine all solutions of the equation: .$\displaystyle 5\sin3x - 11 \:=\:3\sin3x - 12$ . in the domain $\displaystyle 0 \leq x \leq 2\pi$

I've gotten to the point of sin 3x = -0.5,
but I'm not sure of what to do next.
You have: .$\displaystyle \sin3x \:=\:-\tfrac{1}{2}$
You should know that: . $\displaystyle \sin^{-1}\left(\text{-}\tfrac{1}{2}\right) \:=\:\frac{7\pi}{6},\;\frac{11\pi}{6}$ . . . plus its variations.

. . So: .$\displaystyle 3x \;=\;\frac{7\pi}{6},\;\frac{11\pi}{6},\;\frac{19\p i}{6},\;\frac{23\pi}{6},\;\frac{31\pi}{6},\;\frac{ 35\pi}{6}\;\hdots$

Therefore: .$\displaystyle x \;=\;\frac{7\pi}{18},\;\frac{11\pi}{18},\;\frac{19 \pi}{18},\;\frac{23\pi}{18},\;\frac{31\pi}{18},\;\ frac{35\pi}{18}$

4. Thanks to both of you for your help!