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Thread: 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 $\displaystyle 4cos^2x+cosx=0$, let $\displaystyle u=cosx$.
    From that, you will get $\displaystyle 4u^2+u=0$. Factor and solve that for u, then plug $\displaystyle cosx$ 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 $\displaystyle a=\cos x$

    $\displaystyle 4\cos^2x + cosx = 0$

    $\displaystyle 4a^2 + a = 0$

    $\displaystyle a(4a+1) = 0$

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

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

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