Closed form solution of $x + \sin(x) = k$

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?

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.

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 $$\left( \begin{array}{cc} n & x_n \\ 0 & 123.4560000 \\ 1 & 125.4396296 \\ 2 & 124.5477695 \\ 3 & 124.4133695 \\ 4 & 124.4070739 \\ 5 & 124.4070595 \end{array} \right)$$