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Math Help - Help (Trig Equations) (ugh)

  1. #1
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    Help (Trig Equations) (ugh)

    1. 3cot^2 (x) - 1 = 0
    My answer: pi/3, 2pi/3, 4pi/3, 5pi/3

    2. 4cos^2 (x) - 1 = 0
    My answer: pi/3, 2pi/3, 5pi/3, 4pi/3

    3. 2sin (x) + csc (x) = 0
    My answer: unknown lol
    i got to the part: sin^2 (x) = -1/2

    4. 4sin^3 (x) + 2sin^2 (x) - 2sin (x) = 1
    (ahhhhhh)

    Answer verification needed ASAP
    Last edited by NeedHelp18; May 9th 2008 at 01:08 PM.
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  2. #2
    MHF Contributor
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    You're kidding on #4, right? Don't those middle terms cancel each other and leave a much simpler problem?
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  3. #3
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    EDIT: Nvm, sorry.
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  4. #4
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    woops my bad i fixed it lol
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  5. #5
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    Hello, NeedHelp18!

    Ya done good!


    1)\;\;3\cot^2\!x - 1 \:=\: 0
    My answers: . \frac{\pi}{3},\;\frac{2\pi}{3},\;\frac{4\pi}{3},\;  \frac{5\pi}{3} . . . . Right!


    2)\;\;4\cos^2\!x - 1 \:=\: 0
    My answer: . \frac{\pi}{3},\;\frac{2\pi}{3},\;\frac{5\pi}{3},\;  \frac{4\pi}{3} . . . . Yes!


    3)\;\;2\sin x + \csc x \:= \:0
    My answer: unknown ... LOL . . . . Correct!
    i got to the part: . \sin^2\!x \:=\:-\frac{1}{2} . . . . There are no real roots



    4)\;\;4\sin^3\!x + 2\sin^2\!x - 2\sin x \:= \:1
    We have: . 4\sin^3\!x + 2\sin^2\!x - 2\sin x - 1 \;=\;0

    Factor: . 2\sin^2\!x(2\sin x + 1) - (2\sin x + 1) \;=\;0

    Factor: . (2\sin x + 1)(2\sin^2\!x - 1) \;=\;0

    . . \begin{array}{ccccccc}2\sin x +1\:=\:0 & \Rightarrow & \sin x \:=\:-\frac{1}{2} & \Rightarrow & x \:=\:\dfrac{7\pi}{6},\:\dfrac{11\pi}{6} \\ \\[-3mm]<br /> <br />
2\sin^2\!x-1\:=\:0 & \Rightarrow & \sin x \:=\:\pm\frac{1}{\sqrt{2}} & \Rightarrow & x \:=\:\dfrac{\pi}{4},\:\dfrac{3\pi}{4},\:\dfrac{5\p  i}{4},\:\dfrac{7\pi}{4} \end{array}

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  6. #6
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    Sometime i amaze myself
    i was so sure those answers were wrong
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  7. #7
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    . . . . There are no real roots

    Is it because it would be an imaginary number?
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  8. #8
    Moo
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    Quote Originally Posted by NeedHelp18 View Post
    . . . . There are no real roots

    Is it because it would be an imaginary number?
    Yes it is =)
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