$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$ How to find the solution of this trigonometric equation
$$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$$
I have used the formulae $$\cos(x+60^\circ)\cos(x-60^\circ)=\cos^2 60^\circ - \sin^2x$$
$$\sin(x+60^\circ)\sin(x-60^\circ)=\sin^2x-\sin^260^\circ$$
How to move further? What is the least positive value of $x$?

You could convert the following equation to tan and then proceed .By solving you get tan(x+110)=tan3x therefore x = 55 degrees note tan3x is obtained by solving rhs . 
