we can replace tanx with y this making the equation
3y^2 - 5y - 2 = 0 so we solve for y
Δ = b^2 - 4ac (a=3 b=-5 c=-2)
Δ = 25 + 24 = 49
y1 = (5 + 7) / 6 = 12/6 = 2
y2 = (5 - 7) / 6 = -2/6 = -1/3
since tan is positive in the first and third quadrant and according to the question x belongs somewhere in the first two quadrants the following is true
for y1 = tanx = 2 only applies for the first quadrant ( from 0° to 90°) while for y2 = tanx = -1/3 it can only apply for the second quadrant (from 90° to 180°). i hope this made sense because english is not my first language.
