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Math Help - Trig solutions

  1. #1
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    Trig solutions

    a) Sec^2(x) - 2Tan(x) = 4
    b) Cos(x) + 1 = Sin(x)
    c) 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.
    \sec^2\,x - 2\tan\,x = 4

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

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

    Note that the identity
    (1-\cos x)(1+\cos x)=1-\cos^2x=sin^2x
    stands for all possible real numbers x.
    And by squaring up both sides of equation (2), we have
    \sin^2 x=(1+\cos x)(1+\cos x)
    Therefore it's possible to cancel the term \sin^2x:
    (1-\cos x)(1+\cos x)=(1+\cos x)(1+\cos x)\Rightarrow 2\cos x(1+\cos x)=0
    Therefore
    \cos x=0, \sin x=1\;\text{or}\;\cos x=-1, \sin x=0
    That being said,
    x=90^\circ+360^\circ\cdot k\;\text{where $k$ is an integer}
    or
    x=180^\circ+360^\circ\cdot k\;\text{where $k$ is an integer}
    Last edited by Ephemerasylum; August 10th 2010 at 08:12 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.

    2\cos(3x) - 1 = 0

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

    3x=1.0472

    x=.3491

    Of course I get the easy one.
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