Let's first start with the original:
#tan^3(x)-3tan(x)=0# for #[0,2pi)#
For ease sake, let's first make #T=tanx#. We get:
#T^3-3T=0#
We can factor out #T# from both terms on the left and get:
#T(T^2-3)=0#
we can then factor the terms in the brackets. Since there is no #T# term, it means that we're dealing with a factorization of the type #(ax+b)(ax-b)#. We'll get:
#T(T+sqrt3)(T-sqrt3)=0#
#T=tanx=0, -sqrt3, sqrt3#
In finishing solving this, remember that #tan="opp"/"adj"#
#tanx=0#
For #tanx=0#, the opposite needs to be 0. That happens at #0pi, pi# (it happens at #2pi# as well but is outside our range).
#tanx=sqrt3, -sqrt3#
Here we have opposite equal to #sqrt3# and adjacent equal to 1. Because we have both positive and negative values, we can have it where both opposite and adjacent are both positive, both negative, and where only one is. This means that all 4 quadrants have solutions.
We also have a situation where we have a 30, 60, 90 triangle (the ratios of the sides are 1, #sqrt3#, 2). With the opposite being the #sqrt3# side, that means the angle is #60^o#, or #pi/3#.
We have then, #pi/3, (2pi)/3, (4pi)/3, (5pi)/3#
The full answer is, therefore, #x=0, pi/3, (2pi)/3, pi, (4pi)/3, (5pi)/3#
