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.
 A: Actually, the function $f(x)=\sin x + x$ not only grows, but it grows strictly. The function is a continuous injective function on $\mathbb R$, and since $f(0)<1$ and $f(\frac\pi2) > 1$, it is clear that $f(x)=1$ for some $x\in [0,\frac\pi2]$
A: The function is strictly increasing since its derivative  is $>0$ except on the discrete set $\bigl\{\frac\pi2+k\pi,\enspace k\in \mathbf Z\}$. Hence it is a continuous bijection from $\mathbf R$ to $\mathbf R$.
A: 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.
A: 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 $(a,b)$.
A: Let $g:\left[0,\frac{\pi }{2}\right]\rightarrow \mathbb{R},\:g\left(x\right)=f\left(x\right)-1$ where $f(x)=sin(x)+x$ and $g$ is continous on interval. Than $g(0)<0$ and $g\left(\frac{\pi }{2}\right)>0$.
Therefore $g\left(0\right)\cdot g\left(\frac{\pi }{2}\right)<0$ and it has at least a solution. But $g'(x)>0$ on $\left[0,\frac{\pi }{2}\right]$ and it means it is the only solution. Thus $f(x)=1$ has only solution.
The easiest way is to see the graph, where $cos(x)>0$ for any $x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$:

A: By the range of the sine, as$$x=1-\sin x,$$ the root must lie in $(0,2)$.
Now the derivative of $y(x)=\sin x+x-1$ being $\cos x+1$, this function is strictly increasing in this interval. As it is continuous it can have only one root.
And $y(0)=-1<0,y(\pi/2)=\pi/2>0$ brackets this root. A slightly better bracketing is given by $y(1)=\sin(1)>0$, and the first step of the secant method makes a good approximation
$$x\approx\frac1{\sin(1)+1}=0.5430\cdots$$
A: Knowing that the root is between $0$ and $1$ , one can use a few trick to estimate it. For example, a Taylor expansion of $sin(x)$ to third order will give you a cubic equation (for which you can find online solvers). The cubic Taylor expansion absolute error is less than $8 \times 10^{-3}$ for arguments in the our search interval. If you use a Pade approximation:
$\begin{equation}
sin(x) \approx \frac{x-7/60 x^3}{1+1/20 x^2}
\end{equation}$,
the absolute error is less than $2 \times 10^{-4}$. Again, you get a cubic equation, or if you drop the cubic term, you still get a fair estimate ($0.5064$)
Otherwise, doing a fixed point iteration $x = 1-sin(x)$ starting with say $x=0.5$ arrives at the fully converged solution (double precision) in about $200$ iterations: $x_0 = 0.510973429388569$.
Lastly, your problem is a particular case of the famous Kepler's equation (see here), which was extensively studied and solved. So you can use that for your problem.
