# Thread: Finding period of a tan function with x^2 as an exponent

1. ## 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?

2. ## 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.

3. ## Re: Finding period of a tan function with x^2 as an exponent

Originally Posted by GorFree
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 \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*}