Number of solutions of cos5x+cos5(x+2π3)+cos5(x+4π3)=0

Question

amarantha41 · Accepted Answer

Let  f(x)=cos5x+cos5(x+2π3)+cos5(x+4π3)  Because cosine has period 2π, we easily see that f(x) has period 2π3. 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(x)=∑k=05akcos⁡kx  for some constants ak∈R. Period 2π3 implies that ak=0 unless k is divisible by three. At this point we know that  f(x)=a0+a3cos⁡3x  for some constants a0,a3. We have f(x+π)=−f(x) for all x implying that a0=0 (you can see this also by observing that only odd powers of cos⁡x occur in that quintic, or by observing that ∫02πf(x)dx=0)  Last but not least  a3=f(0)=1+(−12)5+(−12)5=1516  So  f(x)=1516cos⁡3x

Alex Sheppard · Accepted Answer

Use cos⁡(A+B) formula to show for a natural number n,  cos⁡nx+cos⁡(nx+2nπ3)+cos⁡(nx+4nπ3)={0when n is coprime to 33cos⁡nxwhen n is a multiple of 3  Combining this with the identity from other answer,  cos5θ=116cos⁡5θ+516cos⁡3θ+1016cos⁡θ  it is easy to see that first and third terms, n=1,5, should sum to zero, while middle term remains to give  cos5x+cos5(x+2π3)+cos5(x+4π3)=1516cos⁡3x

nick1337 · Accepted Answer

Hint: cos5⁡θ=10cos⁡(θ)+5cos⁡(3θ)+cos⁡(5θ)16 because cos5⁡x=132(eix+e−ix)5=e5ix+e−5ix+5(e3ix+e−3ix)+10(eix+e−ix)32 Note that there will be infinitely many solutions as pointed out in comments

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

Answered question

Answer & Explanation

New Questions in Precalculus