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K is a parameter. I tried Wolfram Alpha. With the general parameter, there is no answer at all. With the parameter k = 1, it finds a numerical solution (~ 0.51), but not its interpretation.

Is there any way to solve this equation with a closed form?

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2 Answers 2

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There is no exact analytic solution available to the equation with arbitrary $k$.

For known values of $k$, say, $k=1$ as specified in the question,

$$x+\sin x = 1$$

an approximate close-form solution exists $$x= \frac12\cdot\frac{\pi+2\sqrt3}{3+2\sqrt3}$$

which yields the root 0.51095 vs. the exact 0.51097.

The close-form solution above is derived from a first-order perturbation approximation which turns out to be quite accurate and attractive.

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As already said, there is no closed form and numerical methods are required.

For the case where $k$ is small, you could use a Taylor expansion $$x+\sin(x)=2 x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}+O\left(x^{11 }\right)$$ and use series reversion to get $$x=\frac{k}{2}+\frac{k^3}{96}+\frac{k^5}{1920}+\frac{43 k^7}{1290240}+\frac{223 k^9}{92897280}+O\left(k^{11}\right)$$ which, for $k=1$ would give $$x=\frac{47468023}{92897280}\approx 0.51097323$$ while the "exact" solution would be $0.51097343$.

When $k$ is large, you can notice that $x+\sin(x)$ is bounded by $x\pm 1$ and you could start Newton method with $x_0=k$. Trying for $k=123.456$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 123.4560000 \\ 1 & 125.4396296 \\ 2 & 124.5477695 \\ 3 & 124.4133695 \\ 4 & 124.4070739 \\ 5 & 124.4070595 \end{array} \right)$$

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