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##x=pi/4+(kpi)/2## where ##k## is any integer

Factor using the difference of squares technique.

##a^2-b^2=(a+b)(a-b)##

Thus

##tan^2x-1=(tanx+1)(tanx-1)##

Now substitute this in the original equation.

##(tanx+1)(tanx-1)=0##

When the product of any number of terms is equal to ##0## one of the terms must be equal to ##0## at any time. To solve for ##x## set both multiplied terms equal to ##0##.

##{(tanx+1=0)(tanx-1=0):}##

Solve for both.

##{(tanx=-1)(tanx=1):}##

These are common values. You should have memorized that ##tan(pi/4)=tan((5pi)/4)=1##. You can apply this to other quadrants for the negative version: ##tan((3pi)/4)=tan((7pi)/4)=-1##.

There is no domain restriction in place. Since the tangent function along with the other trigonometric functions is periodic this will have an infinite amount of answers.

Notice that the answers we already found start at ##pi/4## and increase by ##(2pi)/4=pi/2## each time. This will continue to hold true even for things like ##(9pi)/4(17pi)/4-pi/4##.

Thus

##x=pi/4+(kpi)/2## where ##k## is any integer