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Math Help - Solving Trig Equations

  1. #1
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    Solving Trig Equations

    4cos2x = - √8

    cos(3x + 180) = √3/2

    cos2x-sin2xsecx = 1

    Thanks in advance.
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  2. #2
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    What have you tried with these?
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  3. #3
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    Well we're either factoring or using the trig identities (along with addition/half/double angle formulas)
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  4. #4
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    Some identities could be helpful here, but maybe you don't need them at all.

    Take #1 for example

    \displaystyle 4\cos 2x = -\sqrt{8}

    \displaystyle \cos 2x = \frac{-\sqrt{8}}{4}

    \displaystyle \cos 2x = \frac{-2\sqrt{2}}{4}

    \displaystyle \cos 2x = \frac{-\sqrt{2}}{2}

    \displaystyle \cos 2x = \frac{-1}{\sqrt{2}}

    \displaystyle  2x = \frac{3\pi}{4}, \frac{5\pi}{4}

    \displaystyle  x = \frac{3\pi}{8}, \frac{5\pi}{8}
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  5. #5
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    Quote Originally Posted by funkyfiasco View Post
    Well we're either factoring or using the trig identities (along with addition/half/double angle formulas)
    Okay, so for example what have you done with this one?
    4~cos(2x) = - \sqrt{8}

    -Dan

    Edit: Slow on the guns again!
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  6. #6
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    for 4cos2x = -√8 I got 67.5 degrees and 112.5 degrees.

    I'm still trying to find cos(3x + 180 deg) = √3/2 and the other one.

    Thanks for your help
    Last edited by funkyfiasco; March 20th 2011 at 04:56 PM.
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  7. #7
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    Quote Originally Posted by funkyfiasco View Post

    \cos(3x + 180^{o}) = \dfrac{\sqrt{3}}{2}
    .
    Use the addition formula \cos(A+B) = \cos A \cos B - \sin A \sin B

    \cos(3x+180^o) = \cos(3x) \cos(180^o) - \sin(3x) \sin(180^o)

    180 is on the unit circle so you can simplify
    Last edited by e^(i*pi); March 20th 2011 at 04:53 PM. Reason: poor latex
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  8. #8
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    Or you can use the fact that \displaystyle \cos{(\theta + 180^{\circ})} = -\cos{\theta}...
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