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I wanted to ask how you can solve an equation such as:

$$ \sin x = x + 1 $$

How would you solve for x?

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  • $\begingroup$ You'd probably have to do it numerically. $\endgroup$ – user61527 Nov 7 '13 at 4:24
  • $\begingroup$ Not quite a duplicate, although a similar principle should apply. $\endgroup$ – dfeuer Nov 7 '13 at 6:25
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You can close in on an answer, although I doubt if there is a "closed form" expression for it. First, you know sinx oscillates between values of -1 and 1, so the only parts of the line x+1 where there is a solution will be when |x + 1| $\le$ 1. This means you are looking in a domain of [-2,0]. In that domain sinx is negative. It is 0 at x = 0, -1 at x = -$\pi/2$ and again 0 at x = =-$\pi$.

x +1 = -1 where x = -2. sinx already bottomed out at x = -$\pi/2$ , or just past -1.5. So it will cross 1 + x somewhere between -$\pi/2$ and -2, and that is the only solution. If you sketch this you can see that.

Once we get to that point, there is little more analysis that we can do, so as T. Bongers points out, a numerical solution is the most probable approach here. Given that x + 1 and sinx are very nicely behaved functions, and since you know approximately where the answer is, any numerical method that calculates roots of functions should work. You can take a starting point of x = -$\pi/2$, and iterate from there.

If your question is how in general one solves these things, the answer in general is numerically. However, to the extent possible you should analyze the situation similarly to what I sketched out here. The reason is that numerical methods cannot be applied blindly. They have an unpleasant way of getting to the wrong answer if you do not start the first step in the right way.

And if you have no idea what the answer ought to look like, you won't know if it comes out wrong.

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