Cos is nice

Let $a = \cos x$ , $b = \cos \left(x + \dfrac{2\pi}{3}\right)$ and $c = \cos \left(x + \dfrac{4\pi}{3}\right)$ . Then we have:

$\begin{aligned} \color{#3D99F6}{a+b+c} & = \cos x + \cos \left(x + \frac{2\pi}{3}\right) + \cos \left(x + \frac{4\pi}{3}\right) \\ & = \cos x - \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \\ & = \color{#3D99F6}{0} \\ & \\ \color{#3D99F6}{ab + bc + ca} & = \cos x \cos \left(x + \frac{2\pi}{3}\right) + \cos \left(x + \frac{2\pi}{3}\right)\cos \left(x + \frac{4\pi}{3}\right) + \cos \left(x + \frac{4\pi}{3}\right) \cos x \\ & = - \cos x \left( \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \right) + \left( \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \right) \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \right) - \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \right) \cos x \\ & = - \frac{1}{2} \cos^2 x + \frac{\sqrt{3}}{2} \sin x \cos x + \frac{1}{4} \cos^2 x - \frac{3}{4} \sin^2 x - \frac{1}{2} \cos^2 x - \frac{\sqrt{3}}{2} \sin x \cos x \\ & = \color{#3D99F6}{- \frac{3}{4}} \\ & \\ \color{#3D99F6}{abc} & = \cos x \cos \left(x + \frac{2\pi}{3}\right) \cos \left(x + \frac{4\pi}{3}\right) \\ & = \cos x \left( \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \right) \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \right) \\ & = \cos x \left( \frac{1}{4} \cos^2 x - \frac{3}{4} \sin^2 x \right) \\ & = \cos x \left( \cos^2 x - \frac{3}{4} \right) \\ & = \color{#3D99F6}{ \frac{1}{4} \cos (3x) } \end{aligned}$

Now we can find $a^7+b^7+c^7 = \cos^7 x + \cos^7 \left(x + \dfrac{2\pi}{3}\right) + \cos^7 \left(x + \dfrac{4\pi}{3}\right)$ using Newton's sums method as follows:

$\begin{aligned} a^2+b^2+c^2 & = (a+b+c)^2 - 2(ab+bc+ca) = 0 - 2 \left(-\frac{3}{4} \right) = \frac{3}{2} \\ a^3+b^3+c^3 & = (a+b+c) \frac{3}{2} + (ab+bc+ca)(0) + 3 abc = 0 + \frac{3}{4}(0) + 3 \times \frac{1}{4} \cos (3x) = \frac{3}{4} \cos (3x) \\ a^4+b^4+c^4 & =(a+b+c) \frac{3}{4} \cos (3x) + (ab+bc+ca) \frac{3}{2} + abc(0)= 0 + \frac{3}{4}\times \frac{3}{2} + \frac{1}{4} \cos (3x) (0) = \frac{9}{8} \\ a^5+b^5+c^5 & = 0 + \frac{3}{4}\times \frac{3}{4} \cos (3x) + \frac{1}{4} \cos (3x) \times \frac{3}{2} = \frac{15}{16} \cos (3x) \\ a^7+b^7+c^7 & = 0 + \frac{3}{4}\times \frac{15}{16} \cos (3x) + \frac{1}{4} \cos (3x) \times \frac{9}{8} = \frac{63}{64} \cos (3x) \end{aligned}$

Therefore,

$\begin{aligned} \cos^7 x + \cos^7 \left(x + \frac{2\pi}{3}\right) + \cos^7 \left(x + \frac{4\pi}{3}\right) = \frac{63}{64} \cos (3x) = 0 \end{aligned}$

And its roots in $[0, 2\pi ]$ are $\dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{3\pi}{2}, \dfrac{11\pi}{6}$ . The number of roots is $\boxed{6}$ .

First simplify cos(x+2 pi/3) and cos(x+4 pi/3) Then take cos(x)^7 common i.e, cosx^7(1+(-1/2^7)((1+sqrt(3) tanx)^7+ (1-sqrt(3) tanx)^7)). Then use binomial expansion to eliminate odd terms. Finally, by observing the resulting sum and sub stituting tanx =(+-)1/sqrt(3) we find the equation holds true, there exists 4 such values in given interval also if cosx=0 equation still hold true and this happens at two such values in given interval. Therefore a total of 6 values.