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Math Help - please show steps

  1. #1
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    please show steps

    solve:
    cos^2(x) + cos(x) = cos2x
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chopstixdcb View Post
    solve:
    cos^2(x) + cos(x) = cos2x
    Note that \cos(2x)=2\cos^2x-1.

    Therefore, your equation becomes \cos^2x-\cos x-1=0.

    Letting u=\cos x, we end up with the quadratic equation u^2-u-1=0.

    Can you take it from here?
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  3. #3
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    so now, all i have to do is factor it right?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chopstixdcb View Post
    so now, all i have to do is factor it right?
    If possible; Otherwise, resort to quadratic formula.

    Also, remember this is a quadratic in terms of \cos x, so once you have your solutions, you need to solve \cos x=u_1 and \cos x=u_2 to find the solutions that you're seeking.
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  5. #5
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    oh geez...i feel so dumb right now. okay thanks a lot.
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    If possible; Otherwise, resort to quadratic formula.

    Also, remember this is a quadratic in terms of \cos x, so once you have your solutions, you need to solve \cos x=u_1 and \cos x=u_2 to find the solutions that you're seeking.
    umm, if you don't mind, can you show me the final steps. for some reason i got non real solution and the other one is cosx=1
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  7. #7
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    u^2 - u - 1 = 0

    u^2 - u + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 1 = 0

    \left(u - \frac{1}{2}\right)^2 - \frac{5}{4} = 0

    \left(u - \frac{1}{2}\right)^2 = \frac{5}{4}

    u - \frac{1}{2} = \frac{\pm \sqrt{5}}{2}

    u = \frac{1 \pm \sqrt{5}}{2}.


    Note that -1 \leq \cos{x} \leq 1 for all x. Are there any solutions that don't fit?
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