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Math Help - Double angle/sum stuff

  1. #1
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    Double angle/sum stuff

    sin 3x + cos 2x = 0

    I need to solve for x between the interval [0,2pi)

    In the previous chapter there was a problem that wanted us to prove that sin 3x = 3cos^2xsinx-sin^3x so i decided to try using that. However i'm not sure if i'm going down the right path. How would you guys do it?

    Wolfram Alpha also got

    2(pi * n-tan inverse(1+(root 5)+(root 5 +2(root5))) I'm not sure how to get tan out of all that stuff.

    ~anonymous
    Last edited by RIkT; January 7th 2010 at 07:46 AM.
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  2. #2
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    Hello, RIkT!

    Your idea is a good one . . .

    We need two Multiple-Angle Identities: . \begin{array}{ccc}\sin3x &=& 3\sin x - 4\sin^3\!x \\ \cos2x &=& 1 - 2\sin^2\!x \end{array}


    \text{Solve for }x\!:\;\;\sin 3x + \cos 2x \:=\: 0\;\;\text{ on }[0,\:2\pi]

    We are given: . \sin x + \cos2x \:=\:0

    Apply identities: . (3\sin x - 4\sin^3\!x) + (1 - 2\sin^2\!x) \:=\:0

    And we have: . 4\sin^3\!x + 2\sin^2\!x - 3\sin x - 1 \:=\:0

    . . which factors: . (\sin x + 1)(4\sin^2\!x - 2\sin x - 1) \:=\:0



    We have two equations to solve:


    (a)\;\;\sin x + 1 \;=\:0 \quad\Rightarrow\quad\sin x \:=\:-1 \quad \Rightarrow\quad \boxed{x \:=\:\frac{3\pi}{2}}


    (b)\;\;4\sin^2\!x - 2\sin x - 1 \:=\:0
    . . Quadratic Formula: . \sin x \;=\;\frac{2\pm\sqrt{(\text{-}2)^2 - 4(4)(\text{-}1)}}{2(4)} \;=\;\frac{1 \pm\sqrt{5}}{4}

    . . \text{Then: }\;\sin x \;=\;\frac{1+\sqrt{5}}{4} \quad\Rightarrow\quad x \;=\;\arcsin\left(\frac{1+\sqrt{5}}{4}\right) \quad\Rightarrow\quad\boxed{ x \;=\;\frac{3\pi}{10},\;\frac{7\pi}{10}}

    . . \text{And: }\;\;\sin x \;=\;\frac{1-\sqrt{5}}{4} \quad\Rightarrow\quad x \;=\;\arcsin\left(\frac{1-\sqrt{5}}{4}\right) \quad\Rightarrow\quad \boxed{x \;=\;\frac{11\pi}{10},\;\frac{19\pi}{10}}

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  3. #3
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    Thanks for the help. I got to the cubic equation. I had trouble factoring it. I got it now. Thank you.
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