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Math Help - Solving Trig eqations

  1. #1
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    Solving Trig eqations

    I have two questions:

    1-How do I go about solving the equation 3cot² + 5csc =-1
    (i couldn't get the angle symbol to appear)


    2- I am supposed to solve this eqation (sec(x)=x-3) graphically. I can get the graph just fine but am now confused as to what the answers are.
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  2. #2
    Super Member wingless's Avatar
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    1-
    3 \text{Cot}^2x + 5 \text{Csc} x = -1

    3 \text{Cot}^2x + \frac{5}{\text{Sin} x} + 1 = 0

    \frac{3\text{Cos}^2x}{\text{Sin}^2x} + \frac{5}{\text{Sin} x} + 1 = 0

     \frac{3\text{Cos}^2x + 5 \text{Sin} x + \text{Sin}^2x}{\text{Sin}^2x}= 0

    (for x\neq \text{k$\pi $})

    3\text{Cos}^2x + 5 \text{Sin} x + \text{Sin}^2x= 0

    3\left(1 - \text{Sin}^2x\right) + 5 \text{Sin} x + \text{Sin}^2x = 0

    3 - 2\text{Sin}^2x + 5 \text{Sin} x = 0 (multiply both sides by - 1)

    2\text{Sin}^2x- 5 \text{Sin} x - 3 = 0

    (2\text{Sin} x + 1)(\text{Sin} x -3) = 0

    2\text{Sin} x + 1 =0 < \text{and} > (\text{Sin} x -3) = 0

    As \text{Sin} x -3 can' t be equal to 0

    2\text{Sin} x + 1 =0

    \text{Sin} x =- \frac{1}{2}

    x = \text{ArcSin} -\frac{1}{2}

    x = \frac{11\pi }{6} + 2\text{k$\pi $}


    --------------------------------------

    2- Graph y = Sec(x) and y = x - 3.
    x coordinates of the point(s) they intersect are your x's and the y coordinates of these points are your Sec(x) and x-3.
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  3. #3
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    Hello, Swim321!

    Solve: . 3\cot^2\!\theta + 5\csc\theta \:=\:-1

    We have: . 3(\csc^2\!\theta-1) + 5\csc\theta +1\:=\:0 \quad\Rightarrow\quad 3\csc^2\!\theta - 3 + 5\csc\theta + 1 \:=\:0<br />

    We have the quadratic: . 3\csc^2\!\theta + 5\csc\theta + 3 \:=\:0

    . . which factors: . (3\csc\theta-1)(\csc x + 2) \:=\:0


    Then we have: . 3\csc\theta-1\:=\:0\quad\Rightarrow\quad \csc\theta \:=\:\frac{1}{3} . . . . which has no real roots

    . . and: . \csc\theta +2\:=\:0\quad\Rightarrow\quad\csc\theta \:=\:-2\quad\Rightarrow\quad \theta \:=\:\begin{Bmatrix}\dfrac{7\pi}{6} + 2\pi n \\ \dfrac{11\pi}{6} + 2\pi n\end{Bmatrix}

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