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Math Help - Help with solving trig equations

  1. #1
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    Help with solving trig equations

    I'm really not sure where to start with these, a solution would be perfect but a method would also be an easier point i guess as i could probably solve them if i knew where to start.

    1)  2tan (\frac {x} {2} ) +3tanx = 0

    2)  2sec2x - cot2x = tan2x

    3)  2 cosec2x = 1 + tan^2 x

    4) find the value of sinx and cosx when cos2x = 1/9

    any help is appreciated, its probably simple and i just cant remember the method =/
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) \displaystyle \tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}
    Let \tan\frac{x}{2}=t\Rightarrow 2t\left(1+\frac{3}{1-t^2}\right)=0.
    Then t_1=0, \ t_2=-2, \ t_3=2

    2) \displaystyle\frac{2}{\cos 2x}-\frac{\cos 2x}{\sin 2x}=\frac{\sin 2x}{\cos 2x}
    2\sin 2x=\cos^2 2x+\sin^22x=1\Rightarrow \sin 2x=\frac{1}{2}
    x=(-1)^k\frac{\pi}{12}+\frac{k\pi}{2}, \ k\in\mathbf{Z}

    3) \frac{2}{\sin 2x}=\frac{1}{\cos^2x}\Rightarrow \sin x=\cos x\Rightarrow \tan x=1\Rightarrow x=\frac{\pi}{4}+k\pi

    4) \cos x=\pm\sqrt{\frac{1+\cos 2x}{2}}
    \sin x=\pm\sqrt{\frac{1-\cos 2x}{2}}
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  3. #3
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    Quote Originally Posted by red_dog View Post
    1) \displaystyle \tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}
    Let \tan\frac{x}{2}=t\Rightarrow 2t\left(1+\frac{3}{1-t^2}\right)=0.
    Then t_1=0, \ t_2=-2, \ t_3=2

    2) \displaystyle\frac{2}{\cos 2x}-\frac{\cos 2x}{\sin 2x}=\frac{\sin 2x}{\cos 2x}
    2\sin 2x=\cos^2 2x+\sin^22x=1\Rightarrow \sin 2x=\frac{1}{2}
    x=(-1)^k\frac{\pi}{12}+\frac{k\pi}{2}, \ k\in\mathbf{Z}

    3) \frac{2}{\sin 2x}=\frac{1}{\cos^2x}\Rightarrow \sin x=\cos x\Rightarrow \tan x=1\Rightarrow x=\frac{\pi}{4}+k\pi

    4) \cos x=\pm\sqrt{\frac{1+\cos 2x}{2}}
    \sin x=\pm\sqrt{\frac{1-\cos 2x}{2}}

    Thanks for the help, sorry if im asking for too much but could you explain 1 a bit? im not sure i get how you get from let tan x/2 = t to the next step
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