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Math Help - Due tomorrow - solving trigonometric equations

  1. #1
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    Due tomorrow - solving trigonometric equations

    First time posting, hey guys!
    I need urgent h/w help that's due tomorrow.

    Here are the problems:

    2cosxcscx - 4cosx - cscx + 2 (find all the answers in the intervals of 0 to 2pi)

    5sin^23x + 8sin3x - 4
    (i finally was able to factor this and then i don't know how to solve the equations from there)

    sec^2xtan^2x + sec^2x = sec^4x
    (verify identity i think)


    Thanks!

    LAST EDIT: i figured out the third one!
    Last edited by littlelisa; April 21st 2008 at 09:49 PM.
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  2. #2
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    Hello, littlelisa!

    Welcome aboard!
    Here's the first one . . .


    Find all answer for: .  0 \,<\,x\,<\,2\pi

    2\cos x\csc x - 4\cos x - \csc x + 2 \:=\:0
    Factor: . 2\cos x(\csc x - 2) - (\csc x - 2) \;=\;0

    Factor: . (\csc x - 2)(2\cos x-1) \:=\:0


    And solve the two equations . . .
    . . \csc x - 2 \:=\:0\quad\Rightarrow\quad\csc x \:=\:2\quad\Rightarrow\quad x \:=\:\frac{\pi}{6},\;\frac{5\pi}{6}
    . . 2\cos x - 1 \:=\:0\quad\Rightarrow\quad\cos x \:=\:\frac{1}{2}\quad\Rightarrow\quad x \:=\:\frac{\pi}{6},\:\frac{5\pi}{3}

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  3. #3
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    wow, i didn't even think of doing that, thanks so much! i really appreciate it!
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  4. #4
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    Hello again, littlelisa!

    5\sin^2\!3x + 8\sin3x - 4  \:=\:0 . . Solve for: 0 < x < 2\pi

    Factor: . (\sin3x + 2)(5\sin3x - 2) \:=\:0


    And we have two equations to solve . . .

    . . \sin3x + 2 \:=\:0\quad\Rightarrow\quad \sin3x \:=\:-2\:{\color{red}?} . . . no real roots

    . . 5\sin3x-2 \:=\:0\quad\Rightarrow\quad \sin3x \:=\:\frac{2}{5} \quad\Rightarrow\quad 3x \:=\:\sin^{-1}(0.4)

    . . An inverse trig function has many possible values . . .

    . . . . . 3x \;=\;\begin{Bmatrix}0.42 \\ 2.73 \\ 6.69 \\ 9.01 \\ 12.98 \\ 15.30\end{Bmatrix}

    . . Therefore: . x \;=\;\begin{Bmatrix}0.14 \\ 0.91 \\ 2.23 \\ 3.00 \\ 4.33 \\ 5.10 \end{Bmatrix}
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