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Math Help - Finding period of a tan function with x^2 as an exponent

  1. #1
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    Finding period of a tan function with x^2 as an exponent

    Find four solutions of tan(2x^2 + x − 1) = 5

    I work this out to:

    2x^2 + x -2.373 = 0
    Use the quadratic formula to get x= .8676, -1.3676

    I don't know how to find two other answers since I don't the period.

    If it didn't have x^2 I could put it in the form: A tan((2pi/B)(x-C)+D where B is the period.

    How does one find the other solutions?
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  2. #2
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    Re: Finding period of a tan function with x^2 as an exponent

    As the tangent is positive (+5) there are solutions in the first and third quadrants.
    So inverse tan of 5=1.373 or pi+1.373 or 2pi+1.373 or 3pi+1.373 and so on.
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  3. #3
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    Re: Finding period of a tan function with x^2 as an exponent

    Quote Originally Posted by GorFree View Post
    Find four solutions of tan(2x^2 + x − 1) = 5

    I work this out to:

    2x^2 + x -2.373 = 0
    Use the quadratic formula to get x= .8676, -1.3676

    I don't know how to find two other answers since I don't the period.

    If it didn't have x^2 I could put it in the form: A tan((2pi/B)(x-C)+D where B is the period.

    How does one find the other solutions?
    It's not periodic. But to solve this equation...

    \displaystyle \begin{align*} \tan{ \left( 2x^2 + x - 1 \right) } &= 5 \\ 2x^2 + x - 1 &= \arctan{(5)} + \pi n, n \in \mathbf{Z} \\ 2x^2 + x - 1 - \arctan{(5)} - \pi n &= 0 \\ x &= \frac{-1 \pm \sqrt{1^2 - 4(2)[-1 - \arctan{(5)} - \pi n]}}{2(2)} \\ x &= \frac{-1 \pm \sqrt{1 + 8 + 8\arctan{(5)} + 8\pi n }}{4} \\ x &= \frac{-1 \pm \sqrt{9 + 8\arctan{(5)} + 8\pi n}}{4}  \end{align*}
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