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Math Help - Solving Quandratic Trigonometric Equations

  1. #1
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    Solving Quandratic Trigonometric Equations

    How would I solve these equations?
    1) (2sinx - 1)cosx = 1
    2) (2 cosx + √3)sinx = 0

    I tried to convert them to the same identity, but wasn't able to find a substitute in neither the pythagorean identities or the double angle formulas, and I'm lost as to what to do now. Any suggestions/advice would be appreciated, thanks in advance!
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  2. #2
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    Re: Solving Quandratic Trigonometric Equations

    Hello, misiaizeska!

    [1]\;(2\sin x - 1)\cos x \:=\: 1

    [2]\;(2\cos x + \sqrt{3})\sin x \:=\: 0

    The second one is easy . . . the product is zero.

    We have:
    . . 2\cos x+\sqrt{3} \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\tfrac{\sqrt{3}}{2} \quad\Rightarrow\quad x \:=\:\begin{Bmatrix}\frac{5\pi}{6} + 2\pi n \\ \frac{7\pi}{6} + 2\pi n \end{Bmatrix}
    . . \sin x \:=\:0 \quad\Rightarrow\quad x \:=\:\pi n


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I have an incomplete solution for the first one.


    We have: . 2\sin x\cos x - \cos x \:=\:1 \quad\Rightarrow\quad 2\sin x\cos x \:=\:1 + \cos x


    Square both sides: . . . . . . . 4\sin^2\!x\cos^2\!x \:=\:1 + 2\cos x + \cos^2\!x

    . . . . . . . . . . . . . . . . . 4(1-\cos^2\!x)\cos^2x \:=\:1 + 2\cos x + \cos^2\!x

    . . . . . . . . . . . . . . . . . . 4\cos^2x - 4\cos^4\!x \:=\:1 + 2\cos x + \cos^2\!x

    n . . . . . . . . 4\cos^4\!x- 3\cos^2\!x + 2\cos x + 1 \:=\:0

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


    I found one root: . \cos x + 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-1 \quad\Rightarrow\quad x \:=\:\pi + 2\pi n

    I don't know if there are any others . . .
    Thanks from MarkFL and misiaizeska
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  3. #3
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    Re: Solving Quandratic Trigonometric Equations

    Thank you very much, that was really helpful and I was able to figure it out!
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