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Math Help - Sine/Cosine Equations

  1. #1
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    Sine/Cosine Equations

    I don't understand how to do these following equations:

    Solve for x:

    2(cos x)^2 = cos x for -(pi) < x < (pi)

    cos 2x = 5sinx - 2 for all real x

    I'm more interested in how to do them rather than the answers themselves.

    Thank you very much, Adam
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  2. #2
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    Hi

    2 \cos^2 x = \cos x

    \cos x\: (2 \cos x -1) = 0

    \cos x  = 0 or \cos x = \frac{1}{2}


    \cos 2x = 5 \sin x - 2

    1 - 2 \sin^2 x = 5 \sin x - 2

    2 \sin^2 x + 5 \sin x - 3 = 0

    2 X^2 + 5 X - 3 = 0 with X = \sin x
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  3. #3
    Air
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    Quote Originally Posted by Adam We View Post
    I don't understand how to do these following equations:

    Solve for x:

    2(cos x)^2 = cos x for -(pi) < x < (pi)

    cos 2x = 5sinx - 2 for all real x

    I'm more interested in how to do them rather than the answers themselves.

    Thank you very much, Adam
    2(\cos x)^2 = \cos x \ \ \ \ -\pi<x<\pi

    Let u=\cos x and rearrange to get it into a quadratic equation form.

    2u^2 - u=0

    This is in quadratic equation form as we have polynomial to power of

    2 thus we can factorise and solve for solutions.
    u(2u-1) = 0
    u = 0 or 2u - 1= 0

    So your solutions will occur when \cos x = 0 and \cos x = \frac{1}{2}.

    As \cos x is a periodic function, thus there may be more than one solution of x for between -\pi<x<\pi. One way to find the solution is to plot the graph between -\pi<x<\pi for \cos x and draw a line at \frac{1}{2} and 0 and see where it intercepts the curve. Those would be you solutions.

    __________________
    \cos 2x = 5\sin x - 2

    You need to know the double angle rule. For \cos 2x, there are three versions and you must pick the appropriate one. For your example, we would pick one involving \sin x. This would be \cos 2x = 1 - 2\sin ^2 x.

    \therefore 1 - 2\sin ^2 x = 5\sin x - 2

    Now you can let u=\sin x and solve for solution of the quadratic equation and replace it with your substitution to find the specific values (as done in the first question).
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