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Math Help - Soving equation involving trigo. func.

  1. #1
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    Soving equation involving trigo. func.

    Solve, for 0<x<2\pi, the equation 2cos^2x+cosxsinx-6sin^2x=0

    I tried to divide both sides by cos^2x and obtain
    2+tanx-6tan^2x=0, which can be solved easily.

    I can only divide both sides by cos^2x provided that it is non-zero, right?

    Do I have to check whether cos^2x=0 will give some other solution to the equation??

    Thanks!!
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  2. #2
    MHF Contributor red_dog's Avatar
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    If \cos x=0 then \sin x=0, but \sin x and \cos x can't be 0 simultaneously.
    So you can divide the equation by \cos^2x without loosing any solution.
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  3. #3
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    Hello, acc100jt!

    Another approach . . .


    Solve, for 0 <x<2\pi, the equation 2\cos^2\!x+\cos x\sin x-6\sin^2\!x\:=\:0

    Factor: . (2\cos x - 3\sin x)(\cos x + 2\sin x) \:=\:0


    And solve the two equations . . .

    2\cos x \,- \,3\sin x \;=\;0\quad\Rightarrow\quad3\sin x\:=\:2\cos x\quad\Rightarrow\quad\frac{\sin x}{\cos x} \,=\,\frac{2}{3}
    . . \tan x \,=\,\frac{2}{3}\quad\Rightarrow\quad x \,=\,\tan^{-1}\left(\frac{2}{3}\right)

    \cos x + 2\sin x\;=\;0\quad\Rightarrow\quad 2\sin x \,=\,\text{-}\cos x\quad\Rightarrow\quad\frac{\sin x}{\cos x} \,=\,\text{-}\frac{1}{2}
    . . \tan x \,=\,\text{-}\frac{1}{2}\quad\Rightarrow\quad x \:=\:\tan^{-1}\left(\text{-}\frac{1}{2}\right)

    . . . with the same results.

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