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Math Help - Trig equation

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by DemonX01 View Post
    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}
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  3. #3
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    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}

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  4. #4
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    Thanks to both of you for your help!
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