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

Question

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 $x$?

Answer 1

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

$$\frac{\cot(x+110^\circ)}{\cot x} = \cot(x+60^\circ)\cdot \cot(x-60^\circ)$$

$$\frac{\cos(x+110^\circ)\cdot \sin x}{\sin (x+110^\circ)\cdot \cos x} = \frac{\cos(x+60^\circ)\cdot \cos(x-60^\circ)}{\sin(x+60^\circ)\cdot \sin(x-60^\circ)}$$

Now Using Componendo and Dividendo, We get

$$\frac{\cos(x+110^\circ)\cdot \sin x+\sin(x+110^\circ)\cdot \cos x}{\cos(x+110^\circ)\cdot \sin x-\sin(x+110^\circ)\cdot \cos x} = \frac{\cos(x+60^\circ)\cdot \cos(x-60^\circ)+\sin(x+60^\circ)\cdot \sin(x-60^\circ)}{\cos(x+60^\circ)\cdot \cos(x-60^\circ)-\sin(x+60^\circ)\cdot \sin(x-60^\circ)}$$

$$-\frac{\sin \left[(x+110^\circ)+x\right]}{\sin\left[(x+110^\circ)-x\right]} = \frac{\cos[(x+60^\circ)-(x-60^\circ)]}{\cos[(x+60^\circ)+(x-60^\circ)]}$$

So we get $$-\frac{\sin(2x+110^\circ)}{\sin (110^\circ)} = \frac{\cos (120^\circ)}{\cos (2x)}$$

$$\sin (2x+110^\circ)\cdot \cos (2x) = \frac{1}{2}\sin (110^0)$$

So we get $$2\sin (2x+110^\circ)\cdot \cos (2x) = \sin(110^0)$$

$$\sin(4x+110^\circ)+\sin(110^\circ) = \sin (110^\circ)$$

So we get $$\sin(4x+110^\circ) =0\Rightarrow 4x+110^0 = n\times 180^\circ\;,$$ Where $n\in \mathbb{Z}$

Answer 2

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

Answer 3

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 .