Trig equation

**DemonX01** · March 6th 2010, 09:04 AM

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.

**skeeter** · March 6th 2010, 09:18 AM

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.

$0 \le x \le 2\pi$

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

let $u = 3x$ ...

$0 \le u \le 6\pi$

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

unit circle values ...

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

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

**Soroban** · March 6th 2010, 09:21 AM

Hello, DemonX01!

A variation of skeeter's solution . . .

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

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

You have: . $\sin3x \:=\:-\tfrac{1}{2}$
You should know that: . $\sin^{-1}\left(\text{-}\tfrac{1}{2}\right) \:=\:\frac{7\pi}{6},\;\frac{11\pi}{6}$ . . . plus its variations.

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

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

**DemonX01** · March 6th 2010, 10:36 AM

Thanks to both of you for your help!

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