# How to solve algebraically the equation $x = \frac{1}{2}\cos\left(\frac 2 3 \sin\left(\frac 3 4 x\right)\right) + 1$

How to solve this trigonometric equation $x = \frac 1 2 \cos\left(\frac 2 3 \sin\left(\frac 3 4 x\right)\right) + 1$ ?

The iterative solution seems to be 1.417.

Can anybody suggest an algebraic solution ? Does it really exist ?

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Why on earth do you want to solve that nasty looking thing? I'm pretty sure even $x = \cos x$ doesn't have a closed-form solution, so I wouldn't hold out much hope for a nice solution to your equation... –  Rahul Feb 5 '11 at 10:43
@Rahul. It is related to a contraction mapping problem. –  Vinod Feb 5 '11 at 10:49
I don't know if this helps you, but after I made the above comment I went looking for a proof that the solution to $x = \cos x$ cannot in fact be expressed in closed form. It turns out that, according to Timothy Chow's 1999 article, this is actually still open! It would be implied by Schanuel's conjecture, but that conjecture is not proven. Perhaps you can adapt Chow's argument to reduce your problem to Schanuel's conjecture too. –  Rahul Feb 5 '11 at 11:25
Shanuel would show it is transcendental, I guess, but that does not mean it cannot be expressed in closed form. –  GEdgar Jun 14 '11 at 12:35
In fact, Schanuel implies much more, the algebraic independence of $\pi$ and $e$, which Timothy Chow was able to extend to show that no solution to $x = \cos x$ can be expressed in closed form (as precisely defined in Chow's paper). See Theorem 1 on p. 443 of Chow's paper. A URL for a freely available version is <www-math.mit.edu/~tchow/closedform.pdf>;. –  Dave L. Renfro Jul 14 '11 at 15:59

You can easily get extremely accurate approximations using (for example) Wims function calculator, by searching the root $x^*$ of $$f(x) = x - \bigg[\frac{1}{2}\cos \bigg(\frac{2}{3}\sin \bigg(\frac{3}{4}x\bigg)\bigg) + 1\bigg]$$ (say between $-5$ and $5$). Computing its value up to a mild number of digits gives $$x^* = 1.417520004....$$ While Inverse symbolic calculator does not recognize this, it does lead to the approximation $$x^* \approx \frac{{10 \sin (1)}}{{e^{\exp (\gamma )} }} = 1.417520089... ,$$ where $\gamma$ is Euler's constant ($=0.5772156649...$).