Well, the title says it all. I've tried utilizing the fact that $\sin^3x+\cos^3x=(\sin x+\cos x)(1-\sin x\cos x)$ and then the equation becomes $(\sin x+\cos x)(2-\sin x\cos x)=1+\cos^2x$.

Squaring both sides, I can write everything in function of $y=2x$. $(1+\sin 2x)(16-8\sin 2x+\sin^2 2x)=9+6\cos 2x+\cos^2 2x$. But then, that doesn't seems to help much.

  • $\begingroup$ What happens if you complete the full cube? $\endgroup$ – abiessu Apr 23 '16 at 2:04
  • 1
    $\begingroup$ Is this supposed to be an identity, or an equation to be solved for $x$? I am hoping it is the latter ... . $\endgroup$ – Oscar Lanzi Apr 23 '16 at 2:05
  • $\begingroup$ It's an equation. Completing the full cube I arrived at the second equation in the main post. $\endgroup$ – Gabriel Apr 23 '16 at 2:06
  • $\begingroup$ By inspection, $2\pi k$ is a root for every $k \in \mathbb{Z}$. There are other roots? $\endgroup$ – Gabriel Apr 23 '16 at 2:18
  • 1
    $\begingroup$ Plotting the LHS in Mathematica, the roots in $[0,2\pi)$ appear to be $x=0$ and $x\approx 2-7.5\times 10^{-5}$. The latter can be expressed using the root of a quintic polynomial, but doesn't otherwise appear to simplify. (Specifically, $x=2\tan^{-1}z$ where $z\approx \pi/2$ is the one real root of $2z^5-z^4-6z^2+2z-1=0$). $\endgroup$ – Semiclassical Apr 23 '16 at 2:19

As Semiclassical commented, using the tangent half-angle substitution changes the equation to $$4 t^6-2 t^5-12 t^3+4 t^2-2 t=0$$ Then $t=0$ is a solution. The remaining quintic (which cannot be solved with radicals), shows only one real root $t \approx 1.55728$ which corresponds to $\frac x2 \approx 0.999963$.

The solution can be refined using numerical such as Newton which would generate the following iterates $$t_{n+1}=\frac{8 t_n^5-3 t_n^4-6 t_n^2+1}{10 t_n^4-4 t_n^3-12 t_n+2}$$ Using, from the plot, $t_0=1.5$, the iterates are then $$t_1=1.56508875739645$$ $$t_2=1.55740134320206$$ $$t_3=1.55727968392768$$ $$t_4=1.55727965380275$$ $$t_5=1.55727965380274$$ which is the solution for fifteen significant figures.

  • $\begingroup$ $\tfrac12 x = 0.999963$ is suspiciously close to $\tfrac12 x = 1$. $\endgroup$ – bubba Apr 23 '16 at 7:19
  • $\begingroup$ @bubba. I cannot disagree, for sure ! Cheers. $\endgroup$ – Claude Leibovici Apr 23 '16 at 7:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.