Painevg

2021-12-28

Number of solutions of ${\mathrm{cos}}^{5}x+{\mathrm{cos}}^{5}(x+\frac{2\pi}{3})+{\mathrm{cos}}^{5}(x+\frac{4\pi}{3})=0$

amarantha41

Beginner2021-12-29Added 38 answers

Let

$f\left(x\right)={\mathrm{cos}}^{5}x+{\mathrm{cos}}^{5}(x+\frac{2\pi}{3})+{\mathrm{cos}}^{5}(x+\frac{4\pi}{3})$

Because cosine has period$2\pi$ , we easily see that f(x) has period $\frac{2\pi}{3}$ . It is well behaved (continuously differentiable), so it can be expanded into a Fourier series. The function f(x) is even, f(-x)=f(x) for all x, so the expansion only contains cosine terms. As you observed, it is a quintic polynomial in cosx, so

$f\left(x\right)=\sum _{k=0}^{5}{a}_{k}\mathrm{cos}kx$

for some constants$a}_{k}\in \mathbb{R$ . Period $\frac{2\pi}{3}$ implies that ${a}_{k}=0$ unless k is divisible by three. At this point we know that

$f\left(x\right)={a}_{0}+{a}_{3}\mathrm{cos}3x$

for some constants$a}_{0},{a}_{3$ . We have $f(x+\pi )=-f\left(x\right)$ for all x implying that ${a}_{0}=0$ (you can see this also by observing that only odd powers of $\mathrm{cos}x$ occur in that quintic, or by observing that ${\int}_{0}^{2\pi}f\left(x\right)dx=0)$

Last but not least

$a}_{3}=f\left(0\right)=1+{\left(\frac{-1}{2}\right)}^{5}+{\left(\frac{-1}{2}\right)}^{5}=\frac{15}{16$

So

$f\left(x\right)=\frac{15}{16}\mathrm{cos}3x$

Because cosine has period

for some constants

for some constants

Last but not least

So

Alex Sheppard

Beginner2021-12-30Added 36 answers

Use $\mathrm{cos}(A+B)$ formula to show for a natural number n,

$\mathrm{cos}nx+\mathrm{cos}(nx+\frac{2n\pi}{3})+\mathrm{cos}(nx+\frac{4n\pi}{3})=\{\begin{array}{cc}0& \text{when n is coprime to 3}\\ 3\mathrm{cos}nx& \text{when n is a multiple of 3}\end{array}$

Combining this with the identity from other answer,

${\mathrm{cos}}^{5}\theta =\frac{1}{16}\mathrm{cos}5\theta +\frac{5}{16}\mathrm{cos}3\theta +\frac{10}{16}\mathrm{cos}\theta$

it is easy to see that first and third terms, n=1,5, should sum to zero, while middle term remains to give

${\mathrm{cos}}^{5}x+{\mathrm{cos}}^{5}(x+\frac{2\pi}{3})+{\mathrm{cos}}^{5}(x+\frac{4\pi}{3})=\frac{15}{16}\mathrm{cos}3x$

Combining this with the identity from other answer,

it is easy to see that first and third terms, n=1,5, should sum to zero, while middle term remains to give

nick1337

Expert2022-01-08Added 777 answers

Hint:

because

Note that there will be infinitely many solutions as pointed out in comments

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