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Math Help - Solve trig equation

  1. #1
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    Solve trig equation

    I can't get the correct ans for this question~~

    Solve the equation for 0 deg <= theta <= 360 deg.

    <br />
\sin 5 \theta/ \sin \theta -  \ cos 5 \theta/\cos \theta  = 2<br />
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  2. #2
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    Quote Originally Posted by ling_c_0202
    I can't get the correct ans for this question~~

    Solve the equation for 0 deg <= theta <= 360 deg.

    <br />
\sin 5 \theta/ \sin \theta -  \ cos 5 \theta/\cos \theta  = 2<br />
    Add fractions (remember commom denominator ):
    \frac{\sin 5x \cos x-\sin x\cos 5x}{\sin x\cos x}=2
    But the numerator is expansion of \sin (5x-x)=\sin 4x
    Thus, you have,
    \frac{\sin 4x}{\sin x\cos x}=2
    Multiply both sides by a half to get,
    \frac{\sin 4x}{2 \sin x\cos x}=1
    But, 2\sin x\cos x=\sin 2x and \sin 4x=2\sin 2x\cos 2x (double angles),
    \frac{2\sin 2x\cos 2x}{\sin 2x}=1
    Notice the \sin 2x kills the other \sin 2x thus,
    2\cos 2x=1
    Thus,
    \cos 2x=1/2
    Thus, all the solutions take the form,
    2x=\left\{ \begin{array} {c}\frac{\pi}{3}+2\pi k\\ -\frac{\pi}{3}+2\pi k
    Thus,
    x=\left\{ \begin{array}{c} \frac{\pi}{6}+\pi k\\ -\frac{\pi}{6}+\pi k
    You need them to be such as, 0\leq x\leq 2\pi
    For the example the top solution is true only for k=1 otherwise it gets too big or too small. Substituting k=1 for the top solution we get,
    \frac{\pi}{6}+\pi=\frac{7\pi}{6}
    The bottom solution is true only for k=1,2 otherwise it gets too big or too small. Substituting k=1,2 for the bottom solution we get,
    -\frac{\pi}{6}+\pi=\frac{5\pi}{6}
    -\frac{\pi}{6}+2\pi=\frac{11\pi}{6}
    Thus,
    x=\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}
    Last edited by ThePerfectHacker; April 19th 2006 at 08:54 AM.
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  3. #3
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    but for the step
    <br />
\frac{2\sin 2x\cos 2x}{\sin 2x}=1<br />

    it is not given in the question sin 2x <> 0,

    how come they could kill each other?

    is it possible if i 'lost' any solution?
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  4. #4
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    Quote Originally Posted by ling_c_0202
    but for the step
    <br />
\frac{2\sin 2x\cos 2x}{\sin 2x}=1<br />

    it is not given in the question sin 2x <> 0,

    how come they could kill each other?

    is it possible if i 'lost' any solution?
    Because \frac{\sin 2x}{\sin 2x}=1 for defined values of x
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  5. #5
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    THANK YOU!!
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