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Math Help - VERY HARD Trigonometric Equation!!

  1. #1
    Junior Member Kaloda's Avatar
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    VERY HARD Trigonometric Equation!!

    2\sin{2x}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3=0

    Solve for x.

    Maple gave 60, 120, 150 & -150 degrees as the answers.

    I tried to find a solution but in vain.
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  2. #2
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    Re: VERY HARD Trigonometric Equation!!

    1. Use an identity to deal with sin(2x).
    2. Use as much algebra as you can.
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  3. #3
    Junior Member Kaloda's Avatar
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    Re: VERY HARD Trigonometric Equation!!

    Quote Originally Posted by phys251 View Post
    1. Use an identity to deal with sin(2x).
    2. Use as much algebra as you can.
    1. I did already.
    2. " ... as much Algebra as you can." = infinity ...
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  4. #4
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    Re: VERY HARD Trigonometric Equation!!

    After using an identity for sin(2x) bring all the terms with 2\sqrt{3} to one side of the equals sign and then square both sides. The identity sin^2x+cos^2x=1 will be helpful later.
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  5. #5
    Junior Member Kaloda's Avatar
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    Re: VERY HARD Trigonometric Equation!!

    haha. THanks. Got it ... I didn't tried that one bcoz I proceeded to more complex solution quickly.
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  6. #6
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    Re: VERY HARD Trigonometric Equation!!

    Hello, Kaloda!

    \text{Solve for }x\!:\;2\sin{2x}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3\:=\:0

    We have: . 2\underbrace{\sin{2x}}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3\:=\:0
    Then: . 2\overbrace{(2\sin x\cos x)} + 2\sqrt{3}\sin x - 2\sqrt{3}\cos x - 3 \:=\:0

    . . . . . . 4\sin x\cos x + 2\sqrt{3}\sin x - 2\sqrt{3}\cos x - 3 \:=\:0

    Factor: . 2\sin x\left(2\cos x + \sqrt{3}\right) - \sqrt{3}\left(2\cos x + \sqrt{3}\right) \:=\:0

    Factor: . \left(2\cos x + \sqrt{3}\right)\left(2\sin x - \sqrt{3}\right) \:=\:0


    Therefore:

    . . 2\cos x + \sqrt{3} \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\frac{\sqrt{3}}{2} \quad\Rightarrow\quad x \:=\:150^o,\,210^o

    . . 2\sin x - \sqrt{3} \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\frac{\sqrt{3}}{2} \quad\Rightarrow\quad x \:=\:60^o,\,120^o
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