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.

Derivative of the LHS is always greater than or equal to zero and it is zero only at points that $\cos(x) = -1$ which is $x = 2n\pi+\pi$. So, the function is always increasing which crosses level $1$ at only one point. Since $\sin (x) + x$ is more than $1$ at point $\frac{\pi}{2}$ and $0$ at point $0$, it should have only one answer in this interval.
