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How do I solve this?$$\frac{1}{\sin(x)} - \frac{1}{\sin(2x)} = \frac{1}{\sin(4x)}$$
I am not sure on how to solve this. This was my attempt:

  1. $ \frac{(2\cos(x)-1)}{(\sin(2x))} = \frac{1}{(2\sin(2x)\cos(2x))}$
  2. $2\cos(x)-1 =\frac{1}{(2\cos(2x))}$
  3. Let $\alpha = \cos(x):2\alpha-1=\frac{1}{(2(2\alpha^2 -1))} $
  4. $8\alpha^3 -4\alpha^2 -4\alpha +1 =0. $
  5. $x= \pm \arccos(\alpha_{i}) +2\pi n, n \in \mathbb{Z} ,i \in \{1,2,3\}$, where $\alpha_i$ = $i$'th root of cubic on step $(4)$

The problem with this is that there is no way to find the roots of that cubic via any method aside from the cubic formula which is not at all recommended. The only thing i know about this cubic is that its 3 roots are real. How do I solve this cubic? Or is this method doomed and there is another method in the trig manipulation that leads to cleaner answers?

Thomas Andrews
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Papa Dhoria
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    Letting $\beta=2\alpha,$ you can simplify the polynomial to,$$\beta^3-\beta^2-2\beta+1=0.$$ Not sure if that helps, it just seems to make it a little less noisy. – Thomas Andrews May 12 '25 at 00:15
  • @ThomasAndrews very similar to the cubic for $2 \cos 2 \pi /7.$ Are we sure about the $\pm$ signs ? – Will Jagy May 12 '25 at 00:18
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    @Papa Dhoria Welcome to the site. Please check this post: https://math.stackexchange.com/questions/2139075/prove-that-cos-pi-7-is-root-of-equation-8x3-4x2-4x1-0 – User May 12 '25 at 00:22
  • @WillJagy , what cubic are you referring to when you say "cubic for 2cos2π/7"? – Papa Dhoria May 12 '25 at 00:23
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    @WillJagy The polynomial looks correct. – Thomas Andrews May 12 '25 at 00:32
  • Papa, the version in Reuschle is $t^3 + t^2 - 2t - 1$ https://archive.org/details/tafelncomplexer00unkngoog/page/n21/mode/2up Here are two chapters about the method of Gauss http://zakuski.math.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf – Will Jagy May 12 '25 at 00:33
  • @ThomasAndrews in that case, we may take your $\beta = -t$ to get $-t^3 - t^2 + 2t + 1 = 0$ so $t^3 + t^2 - 2t - 1 = 0$ and $t = 2 \cos \frac{2k \pi}{7} $ for $k = 1,2,4$ – Will Jagy May 12 '25 at 00:39
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    That means $$x=-\cos(2\pi k/7)=\cos(\pi(2k+7)/7)$$ for $k=1,2,4.$ @WillJagy – Thomas Andrews May 12 '25 at 01:01

1 Answers1

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Hint:

Another way:

Clearly, $\sin x\sin2x\sin4x\ne0\ \ \ \ (1)\iff\sin4x\ne0\iff x\ne\dfrac{m\pi}4$ for any integer $m$

$\dfrac1{\sin x}=\dfrac1{\sin2x}+\dfrac1{\sin4x}=\dfrac{?}{\sin2x\sin4x}$ (use this) $$\iff\sin2x\sin4x=\sin3x\sin2x$$

$$\iff\sin4x=\sin3x\text{ as }\sin2x\ne0$$

Now use $\sin y=\sin A, y=n\pi+(-1)^nA$ where $n$ is any integer, but remember $(1)$

lab bhattacharjee
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