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Math Help - Trig equation check

  1. #1
    RAz
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    Trig equation check

    I would like to know if my answer is correct

    Q: Solve this trig equation where x is an element of 0, 360.

    \frac{cos2x-cos5x}{sin2x+sin5x}=2\sqrt{3}

    \frac{-2sin\frac{1}{2}(7x)sin\frac{1}{2}(-3x)}{2sin\frac{1}{2}(7x)cos\frac{1}{2}(-3x)}=2\sqrt{3}

    \frac{-2sin\frac{1}{2}(7x)}{2sin\frac{1}{2}(7x)} cancels out

    leaving -\frac{sin\frac{1}{2}(-3x)}{cos\frac{1}{2}(-3x)}=2\sqrt{3}

    which is tan\frac{3x}{2}=-2\sqrt{3}

    \frac{3x}{2}=106, 286 (286 is out of range, of \frac{3x}{2} so it gets removed)

    3x=112

    \therefore x=37

    Thanks
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  2. #2
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    I might be wrong but..

    I donīt really understand the first step you have done?
    112/3 = 27,\frac{1}{3} (but I guess you know that).

    Do you get 2\sqrt{3} when u substitute 37,3 back?
    I got something different (I might be wrong)..
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  3. #3
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    Hello, RAz!

    Solve, where x \in (0^o,360^o)\!:\;\;\frac{\cos2x-\cos5x}{\sin2x+\sin5x}\:=\:2\sqrt{3}

    We have: . \frac{-2\sin\left(\dfrac{7x}{2}\right)\sin\left(-\dfrac{3x}{2}\right) }{2\sin\left(\dfrac{7x}{2}\right)\cos\left(-\dfrac{3x}{2}\right)} \;=\;2\sqrt{3} \qquad\Rightarrow\qquad  \frac{\sin\left(\dfrac{7x}{2}\right)\sin\left(\dfr  ac{3x}{2}\right)}{\sin\left(\dfrac{7x}{2}\right)\c  os\left(\dfrac{3x}{2}\right)} \;=\;2\sqrt{3}


    . . . . . . . . . \frac{\sin\left(\dfrac{3x}{2}\right)}{\cos\left(\d  frac{3x}{2}\right)} \;=\;2\sqrt{3} \qquad\Rightarrow\qquad \tan\left(\frac{3x}{2}\right) \;=\;2\sqrt{3}


    Then: . \frac{3x}{2} \;\approx\;73.89^o,\;253.89^o,\;433.89^o


    Therefore: . x \;\approx\;49.26^o,\;169.26^o,\;289.26^o

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