What is the value of x that satisfies 4\cos^{2}30^{\circ}+2x\sin30^{\circ}-\cot^{2}30^{\circ}-6\tan15^{\circ}\tan75^{\circ}=0?

Explanation

Substitute the known trigonometric values: \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \sin 30^{\circ} = \frac{1}{2}, \cot 30^{\circ} = \sqrt{3}. Also, \tan 15^{\circ} \tan 75^{\circ} = \tan 15^{\circ} \cot 15^{\circ} = 1. The equation becomes 4(\frac{3}{4}) + 2x(\frac{1}{2}) - 3 - 6(1) = 0. Simplifying yields 3 + x - 3 - 6 = 0 \implies x = 6.

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