Question

Solve the equation: cot² x - (√3 + 1) cot x + √3 = 0; 0 < x < $\frac{\pi}{2}$

Answer 1

This equation looks like a Quadratic equation in cot x.

it can be solved by factorisation

${cot}^{2} x - \sqrt{3} cot \: x - cot \: x + \sqrt{3} = 0 \\ \\ cot \: x(cot \: x - \sqrt{3} ) - 1(cot \: x - \sqrt{3} = 0 \\ \\ (cot \: x - 1)(cot \: x - \sqrt{3} ) = 0 \\ \\ \\ so \\ \\ cot \: x - 1 = 0 \\ \\ cot \: x = 1 \\ \\ x = {cot}^{ - 1} (1) \\ \\ \\ x = {cot}^{ - 1} (cot \: \frac{\pi}{4} ) \\ \\ x = \frac{\pi}{4} \\ \\ or \\ \\ cot \: x - \sqrt{3} = 0 \\ \\ cot \: x = \sqrt{3} \\ \\ x = {cot}^{ - 1} ( \sqrt{3} ) \\ \\ x = {cot}^{ - 1} ( cot \: \frac{\pi}{6} ) \\ \\ \\ x = \frac{\pi}{6}$