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.
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?