Question 439483
{{{(tan(3x))^2=3}}} is a trigonometric equation.Taking the square root of both 

sides of the equation, we have two equivalent equations: 

{{{tan(3x)=sqrt(3)}}},(1) and, {{{tan(3x)=-sqrt(3)}}}, (2).

As you know the function tan(x) is periodic function with period {{{pi}}}.

Solving each of them we have:{{{3x=n*pi+(pi)/3}}}=> {{{x=n*pi+(pi)/9}}} and

{{{x=n*pi+(4*pi)/9}}}, for equation (1).

{{{3x=n*pi-(pi)/3}}} => {{{x=n*pi-(pi)/9}}} and {{{x=n*pi-(2*pi)/9}}}

for equation (2)