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Math Help - Solving Trigonometric Equations

  1. #1
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    Solving Trigonometric Equations

    Solving trigonometric equations

    I need help on solving these problems within specific intervals. I am having trouble solving for x the most. Any help is again appreciated.

    1)

    5-7sin x = 2cos^2 x x = [ -360, 450 degrees ]

    2)

    2csc^2 x = 3cot^2 x 1 x = [ 0, 2pi ]

    This is just a sampling of my homework, but it will help me a ton if someone can go through these correctly.
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  2. #2
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    Hello, Anna!

    1)\;5 - 7\sin x \:= \:2\cos^2x\qquad [-360^o,\,450^o]

    Replace \cos^2x with 1 - \sin^2x

    . . 5 - 7\sin x\:=\:2(1 - \sin^2x)\quad\Rightarrow\quad 5 - 7\sin x \:=\:2 - 2\sin^2x

    We have a quadratic: . 2\sin^2x - 7\sin x + 3 \:=\:0

    . . which factors: . (\sin x - 3)(2\sin x - 1)\:=\:0


    Then we have: . \sin x - 3\:=\:0\quad\Rightarrow\quad \sin x \:=\:3 ... no real root

    And: . 2\sin x - 1\:=\:0\quad\Rightarrow\quad \sin x \,= \,\frac{1}{2}\quad\Rightarrow\quad\boxed{x \:=\:-330^o,\,-210^o,\,30^o,\,150^o,\,390^o}



    2)\;2\csc^2x \:= \:3\cot^2x - 1\qquad [0,\,2\pi ]

    Replace \cos^2x with \csc^2x - 1

    . . 2\csc^2x \:=\:3(\csc^2x - 1) - 1\quad\Rightarrow\quad 2\csc^2x \:=\:3\csc^2x - 3 - 1

    We have: . \csc^2x\:=\:4\quad\Rightarrow\quad \csc x\:=\:\pm2\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{6},\,\frac{5\pi}{6},\,\frac{7\pi}{  6},\,\frac{11\pi}{6}}

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