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Thread: trigonometry Find the solutions

  1. #1
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    trigonometry Find the solutions

    Use the inverse functions where necessary to find all solutions of the equation in the interval [0, 2pi)

    $\displaystyle
    tan^2 x + 3 tan x - 10 = 0

    $

    Could someone help me with the steps to solve the equation?
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  2. #2
    Senior Member vincisonfire's Avatar
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    Let $\displaystyle *u = tan(x) $*
    $\displaystyle
    u^2 x + 3 u - 10 = 0
    $
    You find $\displaystyle u = tan(x) = \frac{-3\pm\sqrt(9-4\cdot1\cdot-10)}{2} = \frac{-3\pm 7}{2} $ using the quadratic root formula.
    $\displaystyle *x = arctan(\frac{-3\pm 7}{2}) $
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  3. #3
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    Notice this is in quadratic equation form.
    We can let $\displaystyle tan x = y$

    $\displaystyle y^2 + 3 y - 10 = 0$
    $\displaystyle (y + 5)(y - 2) = 0$ (factorising)

    $\displaystyle y = -5$ or $\displaystyle y = 2$

    Now solve for $\displaystyle tan x = -5$ and $\displaystyle tan x = 2$

    Visualising a tan graph we can deduce that:

    For $\displaystyle tan x = -5$, there will be a solution $\displaystyle \frac{pi}{2}<x<pi$
    and $\displaystyle \frac{3pi}{2}<x<2pi$

    For $\displaystyle tan x = 2$, there will be a solution $\displaystyle 0<x<\frac{pi}{2}$ and $\displaystyle pi<x<\frac{3pi}{2}$
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  4. #4
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    Cool, I never thought about using the quadratic formula to solve. Anyways, this is what I came up with:

    $\displaystyle
    (tan (x) - 2) 3(tan (x) + 5)
    $

    $\displaystyle
    arctan(2)= 1.1071487

    $

    and

    $\displaystyle

    arctan(-5) = -1.373400767

    $

    But I cannot use -1.37 since it is outside of the scope of [0, 2pi) correct?
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  5. #5
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    I was wrong, there are four solutions to the problem:

    1.1071, 4.2487, 1.7682, 4.9098

    because:

    $\displaystyle
    arctan(2) + n\pi
    $
    and
    $\displaystyle
    arctan(-5) + n\pi
    $
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