This is a problem of quadratic equation, but first of all we need to name:
[tex]w = tan(x)[/tex]
Then, the equation above:
[tex]-3tan^{2}(x) +1=0[/tex]
Is converting into:
[tex]-3w^{2}+1=0[/tex]
Then:
[tex]w= \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]
being a = -3, b = 0, c = 1, so there are two values of w:
[tex]\left \{ {{w_{1} = \frac{ \sqrt{3} }{3} } \atop {w_{2} =-\frac{ \sqrt{3} }{3} }} \right.[/tex]
Given that w = tan(x)
[tex]\left \{ {{tan(x_{1} ) = \frac{ \sqrt{3} }{3} } \atop {tan(x_{2} ) =-\frac{ \sqrt{3} }{3} }} \right.[/tex]
Finally:
[tex]x_{1}= \frac{ \pi }{6}[/tex]
[tex]x_{2}= -\frac{ \pi }{6}[/tex]
