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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.

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  • $\begingroup$ What happens if you complete the full cube? $\endgroup$ – abiessu Apr 23 '16 at 2:04
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ $\tfrac12 x = 0.999963$ 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

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