Question 293273
*[Tex \LARGE \tan^{-1}(2x)=-\frac{\pi}{6}] ... Start with the given equation.



*[Tex \LARGE 2x=\tan\left(-\frac{\pi}{6}\right)] ... Take the tangent of both sides to eliminate the inverse tangent.



*[Tex \LARGE 2x=\frac{\sin\left(-\frac{\pi}{6}\right)}{\cos\left(-\frac{\pi}{6}\right)}] ... Break up the tangent into sine over cosine 



*[Tex \LARGE 2x=\frac{-\frac{1}{2}}{\cos\left(-\frac{\pi}{6}\right)}] ... Evaluate the sine of {{{-pi/6}}} to get {{{-1/2}}}



*[Tex \LARGE 2x=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}] ... Evaluate the cosine of {{{-pi/6}}} to get {{{sqrt(3)/2}}}



*[Tex \LARGE 2x=\left(-\frac{1}{2}\right)\left(\frac{2}{\sqrt{3}}\right)] ... Multiply the first fraction by the reciprocal of the second fraction.



*[Tex \LARGE 2x=-\frac{1}{\sqrt{3}}] ... Multiply and simplify



*[Tex \LARGE 2x=-\frac{\sqrt{3}}{3}] ... Rationalize the denominator of the fraction on the right side.



*[Tex \LARGE x=-\frac{\sqrt{3}}{6}] ... Divide both sides by 2. 



So the solution is {{{x=-sqrt(3)/6}}}



Edit: Take note that {{{-sqrt(3)/6=-1/(2*sqrt(3))}}} (just do a bit of algebraic manipulation)