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Math Help - How to solve the following trig equation

  1. #1
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    How to solve the following trig equation

    r=a sin(x)

    r=a cos(2x)

    How would I solve the following equation I know cos(2x)=2cos^2-1,1-sin^2(x)

    But I am not sure where to begin.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: How to solve the following trig equation

    Set them equal to each other  a sin(x) = a cos(2x) , the a's cancel, so you get  sin(x) = cos(2x) .
    Now,  cos(2x) = 1 - 2*sin^2(x) , so  sin(x) = 1-2 sin^2(x) , so you got  sinx + 2 sin^2(x) = 1 , so  sin(x)[1 + 2sin(x)] = 1 . You can see that at  \frac{3\pi}{2}  sin(\frac{3\pi}{2}) = -1 , so  sin(\frac{3\pi}{2})[1 + 2*sin(\frac{3\pi}{2})] = -1*(1 + 2(-1)) = -1*-1 = 1 .
    So your solution set would be  \frac{3\pi}{2} + 2\pi * k where K is an integer.
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  3. #3
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    Re: How to solve the following trig equation

    a\sin(x) = a\cos(2x)

    \sin{x} = 1 - 2\sin^2{x}

    2\sin^2{x} + \sin{x} - 1 = 0

    (2\sin{x} - 1)(\sin{x} + 1)  = 0

    can you finish?
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  4. #4
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    Re: How to solve the following trig equation

    Let's not forget, you can NOT divide by 0. What should have been done is...

    \displaystyle \begin{align*} a\sin{(x)} &= a\cos{(2x)} \\ a\sin{(x)} - a\cos{(2x)} &= 0 \\ a\left[ \sin{(x)} - \cos{(2x)} \right] &= 0 \\ a = 0 \textrm{ or } \sin{(x)} - \cos{(2x)} &= 0 \end{align*}

    and then follow the abovementioned methods...
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  5. #5
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    Re: How to solve the following trig equation

    thanks for your responses. They have been most helpful.
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