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Math Help - how to solve this equation ?

  1. #1
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    how to solve this equation ?

    Hi Guys,

    I am trying to investigate a function, during the process I need to solve this equation:

    sin(x)+cos(2x)=0

    in the region 0<=x<=2*Pai

    How do I solve it ?

    thanks !
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  2. #2
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    Use the identity \displaystyle \cos{2x} \equiv 1 - 2\sin^2{x} to turn your equation into a quadratic, which you can then solve.
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  3. #3
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    Start by writing it as \cos{2x} = -\sin{x} and recall that \cos\left(\varphi+\frac{\pi}{2}\right) = -\sin{\varphi}.
    So we have \cos{2x} = \cos\left(x+\frac{\pi}{2}\right). You should be able to solve it with ease now.
    Hint: if \cos{\theta} = \cos{\alpha} then \theta = 2n\pi\pm \alpha (where n is an integer, or zero).
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  4. #4
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    I used the quadratic idea, I did t=sin(x)
    and got t1=-0.5 and t2=1
    thus, I got:
    sin(x)=-0.5 and sin(x)=1 my solution is x=pai/2 and x=11/6*pai

    but there should be another one at x=3.63, I have no idea why....
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  5. #5
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    Quote Originally Posted by WeeG View Post
    I used the quadratic idea, I did t=sin(x)
    and got t1=-0.5 and t2=1
    thus, I got:
    sin(x)=-0.5 and sin(x)=1 my solution is x=pai/2 and x=11/6*pai

    but there should be another one at x=3.63, I have no idea why....
    \sin{x} = -\dfrac{1}{2} at x = \dfrac{7\pi}{6} and x = \dfrac{11\pi}{6}
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