$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$ 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 $x$?

Apply that on the RHS and put the values of known angles such as $\cos^2 60$. On the LHS convert cot into cos/sin and apply componendo and dividendo on both sides. In LHS you will get terms like $\cos(x+110) + \sin(x+110)$ in the numerator. Convert $\cos (x+110)$ into $\sin (90-(x+110)$. Do this similarly in denominator and apply 
$$\sin(c) + \sin(d) = 2\sin(c+d/2)\cos(c-d/2)$$
And
 $$\sin(c) -\sin(d) = 2\sin(c-d/2)\cos(c+d/2)$$
Your LHS would get simplified into known angles then and you can solve for $x$ later
