MHB Sum of Cosine Values for $x$ | POTW #480 8/24/2021 | $100^{\circ}<x<200^{\circ}$

[anemone]

Here is this week's POTW:

-----

Find the sum of the values of $x$ such that $\cos^3 3x+\cos^3 5x=8\cos^3 4x\cos^3 x$, where $100^{\circ}<x<200^{\circ}$.

-----

 

[anemone]

Congratulations to Opalg for his correct solution(Cool), which you can find below:

Spoiler

By a standard trig formula, $2\cos4x\cos x = \cos5x + \cos3x$. So the given equation becomes $$\begin{aligned}\cos^33x + \cos^35x = (\cos5x + \cos3x)^3 &= \cos^35x + \cos^33x + 3\cos3x\cos5x(\cos3x + \cos5x) \\ &= \cos^35x + \cos^33x + \tfrac32\cos3x\cos5x\cos4x\cos x.\end{aligned}$$ In other words, $$\cos3x\cos5x\cos4x\cos x = 0.$$ But the condition for $\cos nx = 0$ is that $x$ should be an odd multiple of $(90/n)^\circ$. Therefore, in the interval $100^\circ < x < 200^\circ$,

$\cos3x = 0$ when $x= 150^\circ,$
$\cos5x = 0$ when $x= 126^\circ,\, 162^\circ$ or $198^\circ,$
$\cos4x = 0$ when $x= 112.5^\circ$ or $x = 157.5^\circ,$
$\cos x$ is not zero anywhere in that interval.

So the values of $x$ in the interval $100^\circ < x < 200^\circ$ with $\cos^33x + \cos^35x = (\cos5x + \cos3x)^3$ are $112.5^\circ,\,126^\circ,\, 150^\circ,\, 157.5^\circ,\, 162^\circ,\, 198^\circ,$ with sum $906^\circ.$