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Math Help - trig

  1. #1
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    trig

    4cos²x + cosx = 0



    and



    2sin²x - sinx - 1 = 0


    thanks
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by gracey View Post
    4cos²x + cosx = 0



    and



    2sin²x - sinx - 1 = 0


    thanks
    Think of these as "polynomials".

    For 4cos^2x+cosx=0, let u=cosx.
    From that, you will get 4u^2+u=0. Factor and solve that for u, then plug cosx back in for u.

    Take the same approach for the second problem.
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  3. #3
    Super Member wingless's Avatar
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    First convert them to quadratic functions.

    Let a=\cos x

    4\cos^2x + cosx = 0

    4a^2 + a = 0

    a(4a+1) = 0

    1. a=0
    2. 4a+1=0, a=-\frac{1}{4}

    a=0
    \cos x = 0
    \boxed{~x = \frac{\pi}{2} + k\pi~}

    a=-\frac{1}{4}
    \cos x = -\frac{1}{4}
    \boxed{~x=-\arccos \left(-\frac{1}{4}\right)+2 k \pi}
    \boxed{~x=\arccos \left(-\frac{1}{4}\right)+2 k \pi}
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