I am supposed to use Banach's fix-point-theorem to prove that there exists exactly one continous function $g:[0,1] \rightarrow \mathbb{R}$ with $g(x) = x+ \frac{1}{2} \cdot \sin(g(x))$ for every $x \in [0,1]$.
The first thing I need to do is verify that I am allowed to use Banachs's theorem. Hence I need to show that $\text{Im}(g) \subseteq [0,1]$ However, I am not able to do that. I know that for every $x$ we have: $-1 \leq \sin(x) \leq 1$. Therefore $$ x+ \frac{1}{2} \cdot \sin(g(x)) \leq 1 + \frac{1}{2} \cdot 1 = \frac{3}{2}$$ which is obviously more than $1$. Also: $$ x+ \frac{1}{2} \cdot \sin(g(x)) \geq 0 + \frac{1}{2} \cdot (-1) = -\frac{1}{2}$$ I also tried finding $g$ but failed. Any hints on this?