Solve sec² x - 1 = √3 tan x on [0,2π)

Please explain how to solve this with steps because I can't figure it out and the book example doesn't help either.
Thank you.

By pythagorean trig identity we have sec^2(x) - 1 = tan^2(x)

sec^2(x) - 1 = sqrt(3) tan(x)
tan^2(x) = sqrt(3)tan(x)

divide both sides by tan(x)

tan(x) = sqrt(3)

take inverse tangent of x

x = tan^-1(sqrt(3))
x = pi/3

tan^2 x = √3 tan x
tanx (tanx - √3 ) = 0 ----> ...easy to solve now