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Thread: Please help me (solve trigonometry equation)

  1. #1
    Ccrow
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    Please help me (solve trigonometry equation)

    Code:
    Find all solutions of
    2 cos(3x)^2 + sin(3x)
    I can simplify it all the way down to cos(6x) + sin(3x) = 0, however I get stuck here
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ccrow View Post
    Code:
    Find all solutions of
    2 cos(3x)^2 + sin(3x)
    I can simplify it all the way down to cos(6x) + sin(3x) = 0, however I get stuck here
    I don't believe your expression is correct. Try it this way:
    $\displaystyle sin^2(3x) + cos^2(3x) = 1$

    So
    $\displaystyle cos^2(3x) = 1 - sin^2(3x)$

    So your equation becomes:
    $\displaystyle 2 ( 1 - sin^2(3x) ) + sin(3x) = 0$

    $\displaystyle 2sin^2(3x) - sin(3x) - 2 = 0$

    which you can treat as a standard quadratic.

    -Dan
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  3. #3
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    Hello, Ccrow!

    I'll take a guess at what you meant . . .


    Find all solutions of: .$\displaystyle 2\cos^2(3x) + \sin(3x)\:=\:0$

    Replace $\displaystyle \cos^2(3x)$ with $\displaystyle 1 - \sin^2(3x)$

    . . then: .$\displaystyle 2\left[1 - \sin^2(3x)\right] + \sin(3x) \;=\;0$

    And we have the quadratic: .$\displaystyle 2\sin^2(3x) - \sin(3x) - 2 \;=\;0$

    Quadratic Formula: .$\displaystyle \sin(3x) \;= \;\frac{-(-1) \pm\sqrt{(-1)^2 - 4(2)(-2)}}{2(2)} \;= \;\frac{1 \pm\sqrt{17}}{4} $

    We must discard the positive square root: .$\displaystyle \sin(3x) \,\neq\,1.28$


    So we have: .$\displaystyle \sin(3x) = -0.7808\quad\Rightarrow\quad 3x = \begin{Bmatrix}-0.8959 + 2\pi n \\ 4.0375 + 2\pi n\end{Bmatrix}$

    Therefore: .$\displaystyle x\:=\:\begin{Bmatrix}-0.2986 + \frac{2\pi}{3}n \\ 1.3458 + \frac{2\pi}{3}n\end{Bmatrix} $


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  4. #4
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    Soroban you can have $\displaystyle \sin(3x)=1.28$ but you will be required to use complex numbers. I assume that he only wanted the real solutions any way, but just thought id point that out.
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