This is for my trig class, thanks for the help! If you could, could you please show all your work? Thanks again!

f(x) = tan

^{2}(x) + (√3 - 1)[tan(x)] - √3 = 0tan

^{2}(x) + √3[tan(x)] - tan(x) - √3 = 0Factor into

[-1 + tan(x)]*[√3 + tan(x)] = 0

which means

[-1 + tan(x)] = 0 and/or [√3 + tan(x)] = 0

Then

tan(x) = 1

tan

^{-1}(1) = pi/4 radiansFor the other equation

[√3 + tan(x)] = 0

tan(x) = -√3

tan

^{-1}(-√3) = -pi/3so that

**x = pi/4 or -pi/3**in the interval [0, 2pi]