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Math Help - Solving a trig multiple angle?

  1. #1
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    Solving a trig multiple angle?

    I have this expression:

    sin^2(3x)=1/4 (reads sine squared three x equals one fourth)

    when i solved it i got sinx= (plus or minus) 1/2 and the x values were from (pi)/18 to 35(pi)/18 with a total of 12 possible x values on the interval (0, 6(pi)) . Could anyone verify for me if this is correct? thanks a lot!
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  2. #2
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    Hello, nascar77!

    I think I got the same answers . . .


    \sin^2(3x)\:=\:\frac{1}{4}

    When i solved it, i got: . \sin 3x\:=\:\pm\tfrac{1}{2}

    and the x values were from \frac{\pi}{18} to \frac{35\pi}{18}

    with a total of 12 possible x-values on the interval (0,\:6\pi) . . . . Did you mean {\color{blue}(0,\:2\pi)} ?

    \text{We have: }\;\sin(3x) \:=\:\pm\frac{1}{2} \quad\Rightarrow\quad 3x \;=\;\begin{Bmatrix}\dfrac{\pi}{6} + 2\pi n \\ \\[-3mm] \dfrac{5\pi}{6} + 2\pi n \\ \\[-3mm] \dfrac{7\pi}{6} + 2\pi n \\ \\[-3mm] \dfrac{11\pi}{6} + 2\pi n \end{Bmatrix}


    \text{Therefore: }\;x \;=\;\begin{Bmatrix}\dfrac{\pi}{18} + \dfrac{2\pi}{3}n \\ \\[-3mm] \dfrac{5\pi}{18} + \dfrac{2\pi}{3}n \\ \\[-3mm] \dfrac{7\pi}{18} + \dfrac{2\pi}{3}n \\ \\[-3mm] \dfrac{11\pi}{18} + \dfrac{2\pi}{3}n  \end{Bmatrix} . = \;\begin{pmatrix}\dfrac{\pi}{18} & \dfrac{13\pi}{18} & \dfrac{25\pi}{18} \\ \\[-3mm] \dfrac{5\pi}{18} & \dfrac{17\pi}{18} & \dfrac{29\pi}{18} \\ \\[-3mm] \dfrac{7\pi}{18} & \dfrac{19\pi}{18} & \dfrac{31\pi}{18} \\ \\[-3mm] \dfrac{11\pi}{18} & \dfrac{23\pi}{18} & \dfrac{35\pi}{18} \end{pmatrix}



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