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  1. #1
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    help

    sin2x=cos3x, solve for x
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  2. #2
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    Quote Originally Posted by lexmex1546 View Post
    sin2x=cos3x, solve for x
    You can do this,
    $\displaystyle \sin 2x-\cos 3x=0$
    $\displaystyle \sin 2x-\sin \left( \frac{\pi}{2} -3x \right)=0$

    Use identity,
    $\displaystyle \sin x-\sin y=2\cos \left( \frac{x+y}{2} \right)\sin \left(\frac{x-y}{2} \right)$

    Thus,
    $\displaystyle 2\cos \left( \frac{\pi}{4}-\frac{5x}{2} \right) \sin \left( \frac{x}{2}-\frac{\pi}{4} \right)=0$

    Thus, you have a two factors equal to zero.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You can do this,
    $\displaystyle \sin 2x-\cos 3x=0$
    $\displaystyle \sin 2x-\sin \left( \frac{\pi}{2} -3x \right)=0$

    Use identity,
    $\displaystyle \sin x-\sin y=2\cos \left( \frac{x+y}{2} \right)\sin \left(\frac{x-y}{2} \right)$

    Thus,
    $\displaystyle 2\cos \left( \frac{\pi}{4}-\frac{5x}{2} \right) \sin \left( \frac{x}{2}-\frac{\pi}{4} \right)=0$

    Thus, you have a two factors equal to zero.
    Oh now that's just neat!

    -Dan
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    Hello, lexmex1546!

    Here's a "cute" way to approach this problem . . .


    Solve for $\displaystyle x:\;\;\sin2x\:=\:\cos3x$

    $\displaystyle \text{Since }\,\sin\theta = \frac{opp}{hyp}\, \text{ and } \,\cos\theta = \frac{adj}{hyp}$

    . . we see that $\displaystyle 2x$ and $\displaystyle 3x$ are angles in the same right triangle . . .


    Code:
                            *
                         *  *
                      *  3x *
                   *        *
                *           *
             *              *
          * 2x              *
       *  *  *  *  *  *  *  *

    Therefore: .$\displaystyle 2x + 3x \:=\:90^o\quad\Rightarrow\quad\boxed{x \,=\,18^o}$

    Of course, this is not a complete solution . . .

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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lexmex1546 View Post
    sin2x=cos3x, solve for x
    The hard way.

    $\displaystyle sin(2x) = 2sin(x)cos(x)$

    $\displaystyle cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)$

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

    $\displaystyle cos(3x) = cos(x) - 4sin^2(x)cos(x)$

    Thus:
    $\displaystyle 2sin(x)cos(x) = cos(x) - 4sin^2(x)cos(x)$

    Pull everything to the LHS and factor the common cos(x):
    $\displaystyle (cos(x))(4sin^2(x) + 2sin(x) - 1) = 0$

    So either $\displaystyle cos(x) = 0$ or $\displaystyle 4sin^2(x) + 2sin(x) - 1 = 0$

    For the sin solution, use the quadratic formula to find:
    $\displaystyle sin(x) = \frac{-1 \pm \sqrt{5}}{4}$

    So $\displaystyle x = asn \left ( \frac{-1 \pm \sqrt{5}}{4} \right )$

    So the whole solution set in the domain $\displaystyle [0, 2\pi)$ will be x = $\displaystyle \pi/2, 3\pi/2$ from the $\displaystyle cos(x) = 0$ solution and x = $\displaystyle asn \left ( \frac{-1 + \sqrt{5}}{4} \right ) = \frac{\pi}{10}, \pi - \frac{\pi}{10} = \frac{9\pi}{10}$ from the sine solution. These may be easily seen to be the same solutions as ThePerfectHacker gave you, only from a much messier derivation.

    -Dan
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