\( \therefore 0 \tan ^ { 2 } 30 ^ { \circ } - \cos ^ { 2 } 60 ^ { \circ } + x \cot ^ { 2 } 30 ^ { \circ } = 3 \)

Let's break it down a bit! First, we know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), so \( \tan^2 30^\circ = \frac{1}{3} \) and \( \cot^2 30^\circ = 3 \). Additionally, \( \cos 60^\circ = \frac{1}{2} \), thus \( \cos^2 60^\circ = \frac{1}{4} \). Now, substituting these values into the equation: \[ 0 \cdot \tan^2 30^\circ - \frac{1}{4} + x \cdot 3 = 3 \] This simplifies to: \[ -x \cdot \frac{1}{4} + 3x = 3 \] Combine the terms and isolate \( x \): \[ 3x - \frac{1}{4} = 3 \] Now just solve for \( x \). Adding \( \frac{1}{4} \) to both sides: \[ 3x = 3 + \frac{1}{4} = \frac{13}{4} \] Therefore, divide by 3: \[ x = \frac{13}{12}. \] So the solution is \( x = \frac{13}{12} \).