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Thread: Solving for 0 ≤ x ≤ 360

  1. #1
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    Solving for 0 ≤ x ≤ 360

    Oh man I suck at these. Can anyone help out please?

    Solve the following equations for 0 ≤ x ≤ 360

    a) $\displaystyle sin2x + sinx = 0$

    b) $\displaystyle 3cos2x + 2 + cosx = 0$

    A kind of method would be fantastic, Thanks guys!
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  2. #2
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    Quote Originally Posted by Nick87 View Post
    Oh man I suck at these. Can anyone help out please?

    Solve the following equations for 0 ≤ x ≤ 360

    a) $\displaystyle sin2x + sinx = 0$
    Remember that $\displaystyle \sin 2x=2 \sin x \cos x$

    $\displaystyle 2 \sin x \cos x+ \sin x=0$

    Factor by $\displaystyle \sin x$ :

    $\displaystyle \sin x (2 \cos x+1)=0$

    So either $\displaystyle \sin x=0$, either $\displaystyle 2 \cos x+1=0$



    b) $\displaystyle 3cos2x + 2 + cosx = 0$
    note that $\displaystyle \cos 2x=2 \cos^2 x-1$.

    Then substitute $\displaystyle t=\cos x$ and solve for t

    A kind of method would be fantastic, Thanks guys!
    We have someone fantastic around here
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    We sure do. Thanks Moo!
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