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Math Help - 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)

    <br />
tan^2 x + 3 tan x - 10 = 0<br /> <br />

    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 *u = tan(x) *
    <br />
u^2 x + 3 u - 10 = 0<br />
    You find  u = tan(x) = \frac{-3\pm\sqrt(9-4\cdot1\cdot-10)}{2} = \frac{-3\pm 7}{2} using the quadratic root formula.
    *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 tan x = y

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

    y = -5 or y = 2

    Now solve for tan x = -5 and tan x = 2

    Visualising a tan graph we can deduce that:

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

    For tan x = 2, there will be a solution 0<x<\frac{pi}{2} and 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:

    <br />
(tan (x) - 2) 3(tan (x) + 5)<br />

    <br />
arctan(2)= 1.1071487    <br /> <br />

    and

    <br /> <br />
arctan(-5) = -1.373400767 <br /> <br />

    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:

    <br />
arctan(2) + n\pi<br />
    and
    <br />
arctan(-5) + n\pi<br />
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