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Thread: Trig solutions

  1. #1
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    Trig solutions

    a) $\displaystyle Sec^2(x) - 2Tan(x) = 4$
    b) $\displaystyle Cos(x) + 1 = Sin(x)$
    c) $\displaystyle 2Cos(3x) - 1 = 0$
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  2. #2
    Senior Member eumyang's Avatar
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    I'll do the first one.
    $\displaystyle \sec^2\,x - 2\tan\,x = 4$

    Use the Pythagorean identity $\displaystyle 1 + \tan^2\,\theta = \sec^2\,\theta$:
    $\displaystyle \begin{aligned}
    1 + \tan^2\,x - 2\tan\,x &= 4 \\
    \tan^2\,x - 2\tan\,x - 3 &= 0
    \end{aligned}$

    Hey, this looks like a quadratic! Not only that, it's factorable! Factor this and set each factor equal to zero:
    $\displaystyle \begin{aligned}
    \tan^2\,x - 2\tan\,x - 3 &= 0 \\
    (\tan\,x - 3)(\tan\,x + 1) &= 0 \\
    \end{aligned}$
    ...
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  3. #3
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    I'll do the second one.
    $\displaystyle \cos x+1=\sin x$

    Note that the identity
    $\displaystyle (1-\cos x)(1+\cos x)=1-\cos^2x=sin^2x$
    stands for all possible real numbers $\displaystyle x$.
    And by squaring up both sides of equation (2), we have
    $\displaystyle \sin^2 x=(1+\cos x)(1+\cos x)$
    Therefore it's possible to cancel the term \sin^2x:
    $\displaystyle (1-\cos x)(1+\cos x)=(1+\cos x)(1+\cos x)\Rightarrow 2\cos x(1+\cos x)=0$
    Therefore
    $\displaystyle \cos x=0, \sin x=1\;\text{or}\;\cos x=-1, \sin x=0$
    That being said,
    $\displaystyle x=90^\circ+360^\circ\cdot k\;\text{where $k$ is an integer} $
    or
    $\displaystyle x=180^\circ+360^\circ\cdot k\;\text{where $k$ is an integer} $
    Last edited by Ephemerasylum; Aug 10th 2010 at 07:12 AM. Reason: Tex syntax corrected.
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  4. #4
    Senior Member eumyang's Avatar
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    You're too late. The OP reposted b) and c) in this thread: http://www.mathhelpforum.com/math-he...ns-153161.html . No fault on your part -- she shouldn't have done that.
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  5. #5
    Member Chokfull's Avatar
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    I'll do the third.

    $\displaystyle 2\cos(3x) - 1 = 0$

    $\displaystyle \cos(3x)=\frac{1}{2}$

    $\displaystyle 3x=1.0472$

    $\displaystyle x=.3491$

    Of course I get the easy one.
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