Question

3cot^(2 )x-1=0 Pls Help

Answer 1

Answer:

$x=\frac{\pi }{6} ,\frac{7\pi }{6}$

Step-by-step explanation:

$3*cot^2x-1=0\\\\3*cot^2x=1\\\\cot^2x=\frac{1}{3} \\\\cot(x)=\sqrt{\frac{1}{3} } \\\\cot(x)=\frac{1}{\sqrt{3} } \\\\\frac{1}{tan(x)}=\frac{1}{\sqrt{3} } \\\\tan(x)=\sqrt{3}$

x=tan^-1($\sqrt{3}$)

It has to be in Quadrant 1  $\frac{+}{+} =+$ or 3 $\frac{-}{-}=+$ for positive.

$x=\frac{\pi }{6} ,\frac{7\pi }{6}$

Unless that incorrect due to tan(x) range is $-\frac{\pi }{2} \leq x\leq \frac{\pi }{2}$. Then the only answer is $x=\frac{\pi }{6}$