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Math Help - goniometric function

  1. #1
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    Prague, Czech Republic
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    goniometric function

    Hello, could you please help me with this:

    I would like to know, if there are some mistakes or absences.

    Thank you so much, M.

    2sin^2x+cos^2x+sinxcosx=1

    2(1-cos^2x)+cos^2x+sinxcosx=1

    2-2cos^2x+cos^2x+sinxcosx=sin^2x+cos^2x

    2-cos^2x+sinxcosx=sin^2x+1-sin^2x

    2-(1-sin^2x)+sinxcosx=1

    2-1+sin^2x+sinxcosx=1

    1+sin^2(x)+sinxcosx=1

    sin^2x+sinxcosx+0

    sinx(sinx+cosx)=0

    1. CASE

     sinx=0

    x=k*pi

    2. CASE

    sinx+cosx=0

     \frac{sinx}{cosx}+1=0

     \frac{sinx}{cosx}=-1

     tanx=-1


     x_1=pi/4+k*pi
     x_2=3/4pi+k*pi
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  2. #2
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    Re: goniometric function

    solved thx
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  3. #3
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    Re: goniometric function

    Hello, majovski!

    There's a small error at the end: x_1 \,=\,-\tfrac{\pi}{4} + k\pi


    2\sin^2\!x+\cos^2\!x+\sin x\cos x\:=\:1

    I would solve it like this . . .

    \sin^2\!x + \sin^2\!x + \cos^2x + \sin x\cos x \;=\;1

    \sin^2\!x + 1 + \sin x\cos x \;=\;1

    \sin^2\!x + \sin x\cos x \;=\;0

    \sin x(\sin x + \cos x) \;=\;0


    [1]\;\sin x \,=\,0 \quad\Rightarrow\quad x \,=\,k\pi

    [2]\;\sin x\,+\cos x \,=\,0 \quad\Rightarrow\quad \tan x \,=\,-1 \quad\Rightarrow\quad x \,=\,-\tfrac{\pi}{4} + k\pi
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  4. #4
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    Re: goniometric function

    hello and thanks a lot :-)
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