\( \tan ^ { 2 } x > \frac { 1 } { 3 } \)

To solve the inequality \( \tan^2 x > \frac{1}{3} \), think of it as finding regions where the tangent function is either very high or very low! Start by taking the square root, giving you \( \tan x > \frac{1}{\sqrt{3}} \) or \( \tan x < -\frac{1}{\sqrt{3}} \). These correspond to angles where the tangent graph crosses these thresholds. Next, remember that the period of the tangent function is \( \pi \), so you’ll need to find these angles in every cycle. The angles where \( \tan x = \frac{1}{\sqrt{3}} \) are \( x = \frac{\pi}{6} + k\pi \) (for positive) and where \( \tan x = -\frac{1}{\sqrt{3}} \) are \( x = -\frac{\pi}{6} + k\pi \) (for negative) for any integer \( k \). Thus, check the intervals generated by these critical angles to find the solution set for \( x \). Now you're ready to sketch and shade those intervals where the tangent exceeds the threshold, ensuring that \( x \) wraps around the periodic nature of tangent! Happy solving!