Showing that $\sin(x) + x = 1$ has one, and only one, solution Problem:
Prove that the equation $$\sin(x) + x = 1$$ has one, and only one solution. Additionally, show that this solution exists on the interval $[0, \frac\pi2$]. Then solve the equation for x with an accuracy of 4 digits.
My progress:
I have no problems visualizing the lines of the LHS and the RHS.
Without plotting, I can see that $\sin(x) + x$ would grow, albeit not strictly (it would have its derivative zero at certain points), and being continuous, it would have to cross $y=1$ at least once.
However, my problem lies in the fact that I can see how equation could have either 1, 2, or 3 solutions. And I'm not able to eliminate the possibility of there existing 2 or 3 solutions. This is of course wrong on my part.
Also, I'm not able to prove that the solution must exist on the given interval.
Any help appreciated!
P.S. As far as actually calculating the solution, I'm planning on using Newton's Method which should be a trivial exercise since they've already provided an interval on which the solution exists.

Let $f(x) = \sin x$ and $g(x) = 1-x$.
$f(0) < g(0)$ and $f(\pi/2) > g(\pi/2)$. Since both $f$ and $g$ are continuous functions then there is a point $t\in (0,\pi/2)$ such that $f(t) = g(t)$.
Let $h(x) = \sin x + x -1$
Assume there are two or more solutions, let $a$ and $b$ ($a < b$) be two of them, i.e. $h(a) = h(b) = 0$. Then Rolle's Theorem states that there is a $c\in (a,b)$ so that $h'(c) = 0$. But $h'(x) = \cos x +1$ wich is strictly positive for all $x\in (0,\pi/2)$ wich contradicts the Theorem due to the assumption that there is more than one solution. Note that $(a,b)\subseteq (0,\pi/2)$ so since there is no such $c\in (0,\pi/2)$ then $c$ cannot exist in $(a,b)$.
