How to solve these trigonometry equations? I have to work with the following 5 equations:


*

*$(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)=8$

*$(2\cos 2x+1)(2\cos 6x+1)(2\cos 18x+1)=1$

*$\dfrac{\cos 2x}{\sin 3x}+\dfrac{\cos 6x}{\sin 9x}+\dfrac{\cos 18x}{\sin 27x}=0$

*$\dfrac{\cos x}{\sin 3x}+\dfrac{\cos 3x}{\sin 9x}+\dfrac{\cos 9x}{\sin 27x}=0$

*$\dfrac{1}{\cos x\cos 2x}+\dfrac{1}{\cos 2x\cos 3x}+\dfrac{1}{\cos 3x\cos 4x}=0$
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:


*

*How to solve equations 2, 3, 4, and 5?

*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?


Thank you in advance.
 A: $2. 2\cos2y+1=2(1-2\sin^2y)+1=\dfrac{\sin3y}{\sin y}$ 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 $n=1,2,3$
A: 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
$$\cos \left( x \right)  \left( 32\, \left( \cos \left( x \right) 
 \right) ^{6}-48\, \left( \cos \left( x \right)  \right) ^{4}+22\,
 \left( \cos \left( x \right)  \right) ^{2}-3 \right) 
=0$$
A: 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.
