Solve on the Interval [0,2pi)
tan(x)=1

You have to solve equation

tan2(x)=1.

This equation has many (basically infinite) number of solutions. You have to find onlys those solutions, which belong to interval [0, 2π]

If you take square root from both sides, you will have  two equations:

a) tan(x)=1  b) tan(x)=-1

x=arctan(1)+kπ, k=0, ±1, ±2,..  (tan x isπ is a periodic fucntion and π is its period)

x=π/4+kπ, k=0, ±1, ±2

If k=0, x=π/4∈[0, 2π],

if  k=1, x=π/4+π=5π/4 ∈ [0, 2π]

if  k=2, x=π/4+2π=9π/4 ∉ [0, 2π]

So we have two solutions x=π/4 and x=5π/4 from interval [0, 2π]

Now, in similar way, solve equation

b) tan(x)=-1