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Math Help - Solving trigonometric equations

  1. #1
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    Solving trigonometric equations

    Solve over Real numbers:
    a)5sin^2t+cos^2t-2=0 b)sec^2t+tantt=7 c)root of sint=2sint-1 d)tant-sintcost=0


    Solve over [0, 2pi[ :

    e)sin^2x-cos^2x=0 f)2tan^2x+3tanx-5=0 g)2sin^2x-sinx-1=0 h)6cos^2x+5cosx=0


    solve:
    2(cos(3x-pi)+2=1 "x" element of reals

    if "x" element [0 degrees,180degrees] solve 4sin2(x-5)=1

    Find the exact value:
    cosx=-1/3 pi/2<0<pi

    evaluate
    sin2x tan2x
    Last edited by etheen; April 21st 2009 at 04:30 PM.
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  2. #2
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    Hello, etheen!

    I'll do a few of them . . .


    Solve over Real numbers:

    a)\;5\sin^2\!t+\cos^2\!t-2\:=\:0
    Replace \cos^2\!t with 1 - \sin^2\!t

    5\sin^2\!t + (1 - \sin^2\!t) - 2 \:=\:0 \quad\Rightarrow\quad 4\sin^2\!t - 1 \:=\:0 \quad\Rightarrow\quad 4\sin^2\!t \:=\:1

    . . \sin^2\!t \:=\:\tfrac{1}{4} \quad\Rightarrow\quad \sin t \:=\:\pm\tfrac{1}{2} \quad\Rightarrow\quad t \;=\;\begin{Bmatrix}\frac{\pi}{6} + 2\pi n \\ \\[-3mm] \frac{5\pi}{6} + 2\pi n \end{Bmatrix}




    b)\;\sec^2\!t+\tan t\:=\:7
    Replace \sec^2\!t with \tan^2\!t + 1

    (\tan^2\!t + 1) + \tan t \:=\:7 \quad\Rightarrow\quad \tan^2\!t + \tan t - 6 \:=\:0 \quad\Rightarrow\quad (\tan t - 2)(\tan t + 3) \:=\:0


    We have: . \begin{array}{cccccccccc}<br />
\tan t - 2 \:=\:0 & \Rightarrow & \tan t \:=\:2 & \Rightarrow & t \:=\:\arctan(2) + \pi n \\<br />
\tan t + 3 \:=\:0 & \Rightarrow & \tan t \:=\:\text{-}3 & \Rightarrow & t \:=\:\arctan(\text{-}3) + \pi n \end{array}




    c)\;\sqrt{\sin t}\:=\:2\sin t-1

    Square both sides: . \left(\sqrt{\sin t}\right)^2 \:=\:(2\sin t - 1)^2 \quad\Rightarrow\quad \sin t \:=\:4\sin^2\!t - 4\sin t + 1

    . . 4\sin^2\!t - 5\sin t + 1 \:=\:0 \quad\Rightarrow\quad (\sin t - 1)(4\sin t - 1) \:=\:0


    We have: . \begin{array}{ccccccc}<br />
\sin t - 1 \:=\:0 & \Rightarrow & \sin t \:=\:1 & \Rightarrow & t \:=\:\frac{\pi}{2} + 2\pi n \\ \\[-3mm]<br />
4\sin t - 1 \:=\:0 & \Rightarrow & \sin t \:=\:\frac{1}{4} & \Rightarrow& t \:=\:\arcsin\left(\frac{1}{4}\right) + 2\pi n \end{array}




    d)\;\tan t-\sin t\cos t\:=\:0

    We have: . \frac{\sin t}{\cos t} - \frac{\sin t \cos^2\!t}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin t - \sin t\cos t}{\cos t} \:=\:0

    . . \frac{\sin t(1 - \cos^2\!t)}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin t\cdot\sin^2\!t}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin^3\!t}{\cos t} \:=\:0

    . . \sin^3\!t \:=\:0 \quad\Rightarrow\quad \sin t \:=\:0 \quad\Rightarrow\quad t \:=\:\pi n

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  3. #3
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    hm..i see, I see. These are the harder ones. I'm going to do a few more and ask if I am stuck...Thank you so much :}
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