How to compute a variable that could not be separated into one side of the equation?

Question

For example: $\sin(x)=x+0.5$, how to compute the value of $x$ ?

Answer 1

There is not a general analytical method to solve problems like this. But a numerical method will rapidly yield a $x$ value that satisfies the equation. Rearrange the equation to ... and iterate this \begin{eqnarray*} x_{n+1}= \sin x_n -0.5 \end{eqnarray*} $ x \rightarrow -1.4973 \cdots$.

Answer 2

As already said, numerical methods are required. Plotting the function, you notice that the root is close to $x=-2$.

So, consider the function $$f(x)=x+\frac 12-\sin(x)$$

So, start Newton method using $x_0=-2$; the iterates will be given by $$x_{n+1}=x_n-\frac{x_n-\sin (x_n)+\frac{1}{2}}{1-\cos (x_n)}$$ leading to the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & -2.000000000 \\ 1 & -1.582880420 \\ 2 & -1.500917413 \\ 3 & -1.497307404 \\ 4 & -1.497300389 \end{array} \right)$$

If you notice that the solution is really close to $-\frac \pi 2$, you could have very good approximations building the $[1,n]$ Padé approximants of the function around this point. Let us call $x^{(n)}$ the solution obtained.

They will be $$x^{(1)}=\frac{12+3 \pi -\pi ^2}{2 \pi -14}\approx -1.4974020$$ $$x^{(2)}=\frac{21- \pi ^2}{4 \pi -20}\approx -1.4973030$$ which seem to be acceptable.

Another thing you could do is to let $x=-\frac \pi 2+y$ making the equation to be $$y+\cos (y)-\frac{\pi-1 }{2}=0$$ and use equivalent for $\cos(y)$. Thie approximation would lead to $$y^2-2 y+\pi -3=0 \implies y=1-\sqrt{4-\pi }\implies x=-\frac \pi 2+1-\sqrt{4-\pi }\approx -1.4972991$$