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Thread: 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.

    $\displaystyle
    \sin 5 \theta/ \sin \theta - \ cos 5 \theta/\cos \theta = 2
    $
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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.

    $\displaystyle
    \sin 5 \theta/ \sin \theta - \ cos 5 \theta/\cos \theta = 2
    $
    Add fractions (remember commom denominator ):
    $\displaystyle \frac{\sin 5x \cos x-\sin x\cos 5x}{\sin x\cos x}=2$
    But the numerator is expansion of $\displaystyle \sin (5x-x)=\sin 4x$
    Thus, you have,
    $\displaystyle \frac{\sin 4x}{\sin x\cos x}=2$
    Multiply both sides by a half to get,
    $\displaystyle \frac{\sin 4x}{2 \sin x\cos x}=1$
    But, $\displaystyle 2\sin x\cos x=\sin 2x$ and $\displaystyle \sin 4x=2\sin 2x\cos 2x$ (double angles),
    $\displaystyle \frac{2\sin 2x\cos 2x}{\sin 2x}=1$
    Notice the $\displaystyle \sin 2x$ kills the other $\displaystyle \sin 2x$ thus,
    $\displaystyle 2\cos 2x=1$
    Thus,
    $\displaystyle \cos 2x=1/2$
    Thus, all the solutions take the form,
    $\displaystyle 2x=\left\{ \begin{array} {c}\frac{\pi}{3}+2\pi k\\ -\frac{\pi}{3}+2\pi k$
    Thus,
    $\displaystyle x=\left\{ \begin{array}{c} \frac{\pi}{6}+\pi k\\ -\frac{\pi}{6}+\pi k$
    You need them to be such as, $\displaystyle 0\leq x\leq 2\pi$
    For the example the top solution is true only for $\displaystyle k=1$ otherwise it gets too big or too small. Substituting $\displaystyle k=1$ for the top solution we get,
    $\displaystyle \frac{\pi}{6}+\pi=\frac{7\pi}{6}$
    The bottom solution is true only for $\displaystyle k=1,2$ otherwise it gets too big or too small. Substituting $\displaystyle k=1,2$ for the bottom solution we get,
    $\displaystyle -\frac{\pi}{6}+\pi=\frac{5\pi}{6}$
    $\displaystyle -\frac{\pi}{6}+2\pi=\frac{11\pi}{6}$
    Thus,
    $\displaystyle x=\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}$
    Last edited by ThePerfectHacker; Apr 19th 2006 at 07:54 AM.
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  3. #3
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    but for the step
    $\displaystyle
    \frac{2\sin 2x\cos 2x}{\sin 2x}=1
    $

    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
    $\displaystyle
    \frac{2\sin 2x\cos 2x}{\sin 2x}=1
    $

    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 $\displaystyle \frac{\sin 2x}{\sin 2x}=1$ for defined values of $\displaystyle x$
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  5. #5
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    THANK YOU!!
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