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 ...
$\displaystyle u = \frac{7\pi}{6} \, , \, \frac{11\pi}{6} \, , \, \frac{19\pi}{6} \, , \, \frac{23\pi}{6} \, , \frac{31\pi}{6} \, , \, \frac{35\pi}{6}$
$\displaystyle x = \frac{u}{3} = \frac{7\pi}{18} \, , \, \frac{11\pi}{18} \, , \, \frac{19\pi}{18} \, , \, \frac{23\pi}{18} \, , \frac{31\pi}{18} \, , \, \frac{35\pi}{18}$
Hello, DemonX01!
A variation of skeeter's solution . . .
You have: .$\displaystyle \sin3x \:=\:-\tfrac{1}{2}$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 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: .$\displaystyle x \;=\;\frac{7\pi}{18},\;\frac{11\pi}{18},\;\frac{19 \pi}{18},\;\frac{23\pi}{18},\;\frac{31\pi}{18},\;\ frac{35\pi}{18}$