# How to solve these trigonometry equations?

I have to work with the following 5 equations:

1. $$(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)=8$$
2. $$(2\cos 2x+1)(2\cos 6x+1)(2\cos 18x+1)=1$$
3. $$\dfrac{\cos 2x}{\sin 3x}+\dfrac{\cos 6x}{\sin 9x}+\dfrac{\cos 18x}{\sin 27x}=0$$
4. $$\dfrac{\cos x}{\sin 3x}+\dfrac{\cos 3x}{\sin 9x}+\dfrac{\cos 9x}{\sin 27x}=0$$
5. $$\dfrac{1}{\cos x\cos 2x}+\dfrac{1}{\cos 2x\cos 3x}+\dfrac{1}{\cos 3x\cos 4x}=0$$

These equations have patterns, and I know if we can use the pattern we will solve the equations very easily. I managed to use the pattern on the first equation to find a telescoping series and get this (it is not a full solution but it is the way to solve the first equation):

We have $$\frac{2\tan x}{1-\tan^2x}=\tan 2x$$ therefore

\begin{align*} (1)&\Leftrightarrow\dfrac{1}{(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)}=\dfrac18\\ &\Leftrightarrow\left(2\tan x\cdot\dfrac{1}{1-\tan^2x}\right)\cdot\dfrac{1}{1-\tan^22x}\cdot\dfrac{1}{1-\tan^24x}=\dfrac14\tan x\\ &\Leftrightarrow\left(2\tan 2x\cdot\dfrac{1}{1-\tan^22x}\right)\cdot\dfrac{1}{1-\tan^24x}=\dfrac12\tan x\\ \end{align*}

and so on.

However, I can't manage to solve the last four. I can't find the key equalities like $$\frac{2\tan x}{1-\tan^2x}=\tan 2x$$ in the first equation. So here are my questions:

1. How to solve equations 2, 3, 4, and 5?
2. What is the strategy to find the key equalities like $$\frac{2\tan x}{1-\tan^2x}=\tan 2x$$ to get a telescoping series for each equation?

In the first you can get it by the following way: $$LS=\frac{2\tan{x}}{\tan{2x}}\cdot\frac{2\tan{2x}}{\tan{4x}}\frac{2\tan{4x}}{\tan{8x}}=\frac{8\tan{x}}{\tan{8x}}$$ and the rest is smooth.

$$2. 2\cos2y+1=2(1-2\sin^2y)+1=\dfrac{\sin3y}{\sin y}$$ for $$\sin y\ne0$$

$$3.\dfrac{\cos2y}{\sin3y}=\dfrac{2\cos2y\sin y}{2\sin3y\sin y}=?$$

$$4. \dfrac{\cos x}{\sin3x}=\dfrac{\sin(3x-x)}{2\sin3x\sin x}=?$$

$$5.\sin x=\sin((n+1)x-nx)=?$$

Set $$n=1,2,3$$

Using that $$\tan(2x)=2\,{\frac {\tan \left( x \right) }{1- \left( \tan \left( x \right) \right) ^{2}}}$$ $$\tan(4x)={\frac {4\,\tan \left( x \right) -4\, \left( \tan \left( x \right) \right) ^{3}}{1-6\, \left( \tan \left( x \right) \right) ^{2}+ \left( \tan \left( x \right) \right) ^{4}}}$$ we get for your first equation $$\left( 8\, \left( \cos \left( x \right) \right) ^{3}+4\, \left( \cos \left( x \right) \right) ^{2}-4\,\cos \left( x \right) -1 \right) \left( 8\, \left( \cos \left( x \right) \right) ^{3}-4\, \left( \cos \left( x \right) \right) ^{2}-4\,\cos \left( x \right) + 1 \right) =0$$ for #5: Your equation is equivalent to $$\cos \left( 3\,x \right) \cos \left( 4\,x \right) +\cos \left( 2\,x \right) \cos \left( x \right) +\cos \left( 4\,x \right) \cos \left( x \right) =0$$ and this can be written in the form $$\cos \left( x \right) \left( 32\, \left( \cos \left( x \right) \right) ^{6}-48\, \left( \cos \left( x \right) \right) ^{4}+22\, \left( \cos \left( x \right) \right) ^{2}-3 \right) =0$$