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I have to solve $\cos(x)\cos(2x)\cos(4x)=1/8$.

I can express it for $x$ only with $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\cos(4x)=\cos(2x+2x)$, but it only seems to become a really big expression and I have no clue how to proceed after... Any suggestions?

N. F. Taussig
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Rosalie Landry-Ouellet
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4 Answers4

7

You can use the double angle identity $$ \sin 2x = 2\sin x \cos x $$ By multiplying $\sin x$, $$\begin{align*} \cos(x)\cos(2x)\cos(4x) &= \frac{1}{8} \\ \frac{1}{2} \sin (2x) \cos(2x)\cos(4x) &= \frac{1}{8}\sin x \\ \frac{1}{4} \sin (4x)\cos(4x) &= \frac{1}{8}\sin x \\ \frac{1}{8} \sin (8x) &= \frac{1}{8}\sin x \\ \sin (8x) &= \sin x \end{align*}$$ The last equation will have $16$ roots in $[0,2\pi)$, but $0$ and $\pi$ do not solve the original equation since they are introduced by $\sin x$.

Henricus V.
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    Since you multiply by $\sin(x)$, you introduce extra solutions at $0$ and $\pi$ which solve your bottom equation, but do not solve the original equation. – 2'5 9'2 Mar 14 '16 at 19:47
  • @alex.jordan Thanks. It is fixed. – Henricus V. Mar 14 '16 at 19:47
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    Why 15? In fact it has 16. Then you omit $0$ and $\pi$ so there are 14 solutions to the original equation (enumerated in my answer). I don't think it's obvious to count the solutions up to $\sin(x)=\sin(8x)$, unless there is something I am missing. – 2'5 9'2 Mar 14 '16 at 19:50
  • So since I have to find the solutions for x= [0, π/2], which solutions do I consider? – Rosalie Landry-Ouellet Mar 14 '16 at 20:06
  • @Rosalie Landry-Ouellet You only need to check that $0,\pi,2\pi$ are real solutions. You don't need to check the other ones. – Henricus V. Mar 14 '16 at 20:07
  • Thank you! So I know that 0 isn't a valid solution but how do I find the ones that are valid? – Rosalie Landry-Ouellet Mar 14 '16 at 20:20
  • @RosalieLandry-Ouellet The other ones are shown using algebra as above. $0,\pi,2\pi$ are special since they are introduced by $\sin x$. – Henricus V. Mar 14 '16 at 20:20
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For $x\ne n\pi$ (it's true, because $ \cos{x}\cdot\cos{2x}\cdot\cos{4x}\,\Big|_{x=n\pi} \ne\frac{1}{8}$) $$ \cos{x}\cdot\cos{2x}\cdot\cos{4x}=\frac{\sin{x}\cdot\cos{x}\cdot\cos{2x}\cdot\cos{4x}}{\sin{x}} = \\ =\frac{1}{2}\cdot \frac{\sin{2x}\cdot\cos{2x}\cdot\cos{4x}}{\sin{x}}=\frac{1}{4}\cdot \frac{\sin{4x}\cdot\cos{4x}}{\sin{x}}=\frac{1}{8}\cdot \frac{\sin{8x}}{\sin{x}},$$ therefore, $$\sin{8x}=\sin{x}.$$

M. Strochyk
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Let $z=e^{ix}$. Then your relation says $$\left(z+z^{-1}\right)\left(z^2+z^{-2}\right)\left(z^4+z^{-4}\right)=1$$ That is: $$ \begin{align} z^7+z^5+z^3+z+z^{-1}+z^{-3}+z^{-5}+z^{-7}&=1\\ z^{-7}\frac{z^{16}-1}{z^2-1}&=1&\text{(for $z^2\neq1$)}\\ z^{16}-1&=z^9-z^7\\ z^{16}-z^9+z^7-1&=0\\ \left(z^7-1\right)\left(z^9+1\right)&=0 \end{align}$$

So $z$ is either a $7$th root of $1$ or a $9$th root of $-1$. Which means your $x$ is among $$\left\{\frac{2\pi}{7}, \frac{4\pi}{7},\frac{6\pi}{7},\frac{8\pi}{7},\frac{10\pi}{7},\frac{12\pi}{7},\frac{\pi}{9},\frac{3\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9},\frac{11\pi}{9},\frac{13\pi}{9},\frac{15\pi}{9},\frac{17\pi}{9}\right\}$$ or translates by $2\pi k$. We've left out $0\pi/7$ and $9\pi/9$ since they cause $z^2-1$ to equal $0$, invalidating an earlier step here. You can directly check that these angles do not satisfy the original equation.

2'5 9'2
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0

Not the easiest way, but still a valid method to tackle problems like this. Use the tan-half-angle substitution

$$ \begin{align} x & = 2 \tan^{-1}(t) \\ \cos(x) & = \frac{1-t^2}{1+t^2} \\ \sin(x) & = \frac{2 t}{1+t^2} \end{align} $$

Also we need to expand the $\cos(n x)$ terms to make

$$ \cos(x) \left( 2 \cos^2(x)-1\right) \left(1-8 \sin^2(x) \cos^2(x) \right) = \frac{1}{8} $$

Now change the variables $x \rightarrow t$ for

$$ \left( \frac{1-t^2}{1+t^2} \right) \left( \frac{t^4-6 t^2+1}{(1+t^2)^2} \right) \left( \frac{t^8-38 t^6+70 t^4-28 t^2+1}{(1+t^2)^4} \right) = \frac{1}{8} $$

All this can be factored as

$$ \frac{(1-3 t^2)(3 t^{12}-90t^{10}+705 t^8-1660 t^6+1365 t^4-266 t^2+7)}{8(1+t^2)^7} = 0 $$

(thank you CAS)

This is an even function, are we are only concerned about positive roots. The first immediate root is $t=\frac{1}{\sqrt{3}}$, which is

$$ x_1 = 2 \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = 2 \frac{\pi}{6} =\frac{\pi}{3}$$

The other roots I got where

$$ \begin{align} t_2 & = 0.176326980708464 & x_2 = 0.349065850398863 \\ t_3 & = 0.481574618807528 & x_3 = 0.897597901025654 \\ t_4 & = 1.1917535925942 & x_4 = 1.74532925199432\\ t_5 & = 1.2539603376627 & x_5 = 1.79519580205130 \\ t_6 & =2.74747741945462 & x_6 = 2.44346095279206 \\ t_7 & = 4.38128626753482 & x_7 = 2.69279370307696 \end{align}$$

John Alexiou
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