The equation $\tan x = \tan 2x \tan 4x \tan 8x$

Question

In the question we have the equality $$\tan 6^{\circ} \tan 42^{\circ} = \tan 12^{\circ} \tan 24^{\circ}$$ which is equivalent to $$ \tan 6^{\circ} = \tan 12^{\circ} \tan 24^{\circ} \tan 48^{\circ}$$ This means that the equation $$\tan x = \tan 2x \tan 4x \tan 8x$$ has the solution $x =6^{\circ} = \frac{\pi}{30}$. How to find all the solution of this equation?

Answer 1

$$\tan x=\tan 2x\tan 4x\tan 8x$$

$$\frac{\sin x}{\cos x}=\frac{2 \sin x \cos x}{\cos 2x}\left(\frac{2\sin 4x \sin 8x}{2 \cos 4x \cos 8x}\right) $$

If $\sin x\neq0$ then,

$$\frac{\cos 2x}{2\cos^2 x}=\frac{\cos 4x - \cos 12x}{\cos 4x + \cos 12x}$$

Now,using componendo-dividendo,

$$\frac{\cos 2x+ 2\cos^2x}{2\cos^2 x-\cos 2x}=\frac{\cos 4x}{\cos 12x}$$

$$ 4 \cos^2 x -1 =\frac{1}{4\cos ^2 4x-3}$$

$$(2\cos 2x+1)(4(2\cos^2 2x -1)^2-3)=1$$

These are the solutions of above equation :wolframalpha

can someone solve the last equation by hand and edit the answer$?$

also,note that I cancelled some terms like $\cos 4x$,because $\cos 4x=0$ is not in the domain of the equation

Answer 2

$\sin x=0$ is obviously a solution. We will assume for further solutions that $\sin x\neq 0.$

Now:

$$\begin{align}\sin 2x&=2\sin x\cos x\\\sin 4x&=4\sin x\cos x\cos2x\\ \sin 8x&=2\sin 4x\cos 4x\\&=8\sin x\cos x\cos 2x\cos 4x.\end{align}$$

So:

$$\begin{align}\frac{\sin x}{\cos x}&=\tan x\\&=\tan 2x\tan 4x\tan 8x\\&=\frac{64\sin^3 x\cos^3 x \cos 2x}{\cos 8x} \end{align}$$

Cancelling $\sin x\neq 0$ and multiplying both sides by $\cos x\cos 8x$ you get:

$$\begin{align}\cos 8x &= 64\sin^2x\cos^4x\cos 2x\\ &=64(1-\cos^2 x)\cos^4x (2\cos^2x-1) \end{align}$$

If $u=\cos x$ then:

$$\cos 8x = T_8(n)=128 u^8 - 256 u^6 + 160 u^4 - 32 u^2 + 1$$ where $T_n$ is the Chebyshev polynomial of the first kind, $T_n(\cos x)=\cos nx.$

Together, this means we have the equation:

$$q(u)=256 u^8 - 448 u^6 + 224 u^4 - 32 u^2 + 1=0$$

Wolfram Alpha helps me compute:

$$q(u)=\frac{T_{15}(u)T_1(u)}{T_3(u)T_5(u)}\tag 1$$
(I theorized this equality after playing with the numerical roots of $q.$)

Wolfram verified (1), but it is not really a satisfying answer for “why?”

So, $u=\cos x$ is a root of $q$ if and only if $\cos 15x=0$ but $\cos 5x\neq 0$ and $\cos 3x\neq 0.$

This amounts to $x=\frac{m\pi}{30}$ where $\gcd(m,30)=1.$

So, in $[-\pi,\pi],$ the solutions are $$\begin{align}x=0,&\pm\frac{\pi}{30},\pm\frac{29\pi}{30},\pm\frac{7\pi}{30},\pm\frac{23\pi}{30},\\ &\pm\frac{11\pi}{30},\pm\frac{19\pi}{30},\pm\frac{13\pi}{30},\pm\frac{17\pi}{30} \end{align}$$

So, other than $0,$ these are the “primitive roots” of $\cos(15x)=0,$ or the $x$ such that $e^{ix}$ is a primitive $60$th root of unity.

Answer 3

Factorize the equation as follows \begin{align} & \tan x-\tan 2x \tan 4x \tan 8x \\ =& \frac{\sin x \cos 2x \cos 4x \cos 8x-\cos x \sin 2x \sin 4x \sin 8x } {\cos x \cos 2x \cos 4x \cos 8x }\\ =& \frac{\sin x [\cos 2x \cos 4x \cos 8x-(1+\cos 2x)\sin 4x \sin 8x ]} {\cos x \cos 2x \cos 4x \cos 8x }\\ =& \frac{\tan x (\cos 12x -2\sin 2x \sin 8x )} {\cos 4x \cos 8x }\\ =& \frac{\tan x }{\cos8x} \left( 4 \cos^2 4x-3 -4\sin 2x \sin 4x\right)\\ =& \frac{\tan x }{\cos8x} \left(16\cos^4 2x +8\cos^3 2x -16\cos^2 2x-8\cos 2x +1 \right)\\ =& \frac{16\tan x }{\cos8x} \left(\cos^2(2x-\frac{2\pi}3) - \frac12 \cos(2x-\frac{2\pi}3) -\frac14\right)\\ &\hspace{15mm}\times \left(\cos^2(2x+\frac{2\pi}3) - \frac12 \cos(2x+\frac{2\pi}3) -\frac14\right)\\ =& \frac{16\tan x }{\cos8x} \left(\cos(2x-\frac{2\pi}3) - \cos\frac\pi5\right)\left(\cos(2x-\frac{2\pi}3) - \cos\frac{3\pi}5\right)\\ &\hspace{15mm}\times \left(\cos(2x+\frac{2\pi}3) - \cos\frac\pi5\right)\left(\cos(2x+\frac{2\pi}3) - \cos\frac{3\pi}5\right)\\ \end{align} where $ \cos\frac{\pi}5 +\cos\frac{3\pi}5 =\frac12$ and $ \cos\frac{\pi}5\cos\frac{3\pi}5 =-\frac14$ are recognized. Then, the full set of the solutions are $x=n\pi,\>n\pi \pm \frac{\pi}3\pm \frac\pi{10}$ and $n\pi \pm \frac{\pi}3\pm \frac{3\pi}{10} $, or $$x-n\pi=0, \>\pm \frac\pi{30}, \>\pm \frac{7\pi}{30} , \>\pm \frac{13\pi}{30} , \>\pm \frac{19\pi}{30} $$

Answer 4

Another solution.

Consider the equation $$\tan (2x) \tan (4x) \tan (8x)-\tan(x)=0$$ Let $x=\tan^{-1}(t)$, use the multiple angle formulae and obtain $$\frac{t^9-28 t^7+134 t^5-92 t^3+t}{t^8-28 t^6+70 t^4-28 t^2+1}=0$$ Discrading the trivial $t=0$,we then face a quartic equation in $u=t^2$ $$u^4-28 u^3+134 u^2-92 u+1=0$$ which can be solved with radicals. The solutions are all positive and the smallest is $$u=7-2 \sqrt{5}-2 \sqrt{15-6 \sqrt{5}}\implies t=\sqrt{7-2 \sqrt{5}-2 \sqrt{15-6 \sqrt{5}} }$$ and then $$x=\tan ^{-1}\left(\sqrt{7-2 \sqrt{5}-2 \sqrt{15-6 \sqrt{5}}}\right)$$ At this point, I am stuck; using my 50+ years old pocket calculator $x=0.1047197551$ which is $\frac \pi{30}$.

But, as @Thomas Andrews answered, there are many other solutions.