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Math Help - Real Hard Trigonometry Question ?

  1. #1
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    Real Hard Trigonometry Question ?

    cos2x + √3sin2x = 2 ????????????
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  2. #2
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    Re: Real Hard Trigonometry Question ?

    \sqrt{1- sin^2(2x)}+ \sqrt{3} sin(2x)+ 2
    \sqrt{1- sin^2(2x)}= 2- \sqrt{3}sin(2x)

    Let y= sin(2x). \sqrt{1- y^2}= 2- \sqrt{3}y

    Square both sides to get a quadratic equation for y.
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  3. #3
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    Re: Real Hard Trigonometry Question ?

    Isn't cos2x = 1 - 2sin^2 x ?? So why apply the square root ?

    And I think you meant =2 instead of +2 in the first line.
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  4. #4
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    Re: Real Hard Trigonometry Question ?

    Isn't cos2x = 1 - 2sin^2 x ?? So why apply the square root ?

    And I think you meant =2 instead of +2 in the first line.
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  5. #5
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    Re: Real Hard Trigonometry Question ?

    Yes it is. But that's not what HallsOfIvy did. He applied the Pythagorean Identity. Notice how what he did is still in terms of sin(2x), not sin(x)?
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  6. #6
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    Re: Real Hard Trigonometry Question ?

    Hello, OmarAj96!

    \cos2x + \sqrt{3}\sin2x \:=\: 2

    Divide by 2: . \tfrac{1}{2}\cos2x + \tfrac{\sqrt{3}}{2}\sin2x \:=\:1


    Then we have:

    . . \sin\tfrac{\pi}{6}\cos2x + \cos\tfrac{\pi}{6}\sin2x \:=\:1

    . . . . . . . . . . . \sin(2x + \tfrac{\pi}{6}) \:=\:1

    . . . . . . . . . . . . . . 2x + \tfrac{\pi}{6} \:=\: \tfrac{\pi}{2} + 2\pi n

    . . . . . . . . . . . . . . . . . 2x \:=\:\tfrac{\pi}{3} + 2\pi n

    . . . . . . . . . . . . . . . . . . x \:=\:\tfrac {\pi}{6} + \pi n
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