Answer:
[tex]x=\frac{\pi }{6} ,\frac{7\pi }{6}[/tex]
Step-by-step explanation:
[tex]3*cot^2x-1=0\\\\3*cot^2x=1\\\\cot^2x=\frac{1}{3} \\\\cot(x)=\sqrt{\frac{1}{3} } \\\\cot(x)=\frac{1}{\sqrt{3} } \\\\\frac{1}{tan(x)}=\frac{1}{\sqrt{3} } \\\\tan(x)=\sqrt{3} \\[/tex]
x=tan^-1([tex]\sqrt{3}[/tex])
It has to be in Quadrant 1 [tex]\frac{+}{+} =+[/tex] or 3 [tex]\frac{-}{-}=+[/tex] for positive.
[tex]x=\frac{\pi }{6} ,\frac{7\pi }{6}[/tex]
Unless that incorrect due to tan(x) range is [tex]-\frac{\pi }{2} \leq x\leq \frac{\pi }{2}[/tex]. Then the only answer is [tex]x=\frac{\pi }{6}[/tex]
