Accuracy of solution? I have a question that asks me to get the solution to the equation $x+\arcsin(x)={\pi\over 2}$ by using a calculator. (Repeatedly pressing cos?) Then it asks to justify the accuracy of the answer. What does this mean? What am I supposed to do? Please help!

For me it seem to be an exercise on the Contraction Mapping Theorem: rewrite the equation as
$$
x = \pi/2-\arcsin x
$$
so rhs is defined only on $[-1,1]$ and takes values on $[0,\pi]$. By monotone arguments we obtain that there is only one solution of the equation. Now, rewrite it as
$$
\arcsin x = \frac\pi2-x
$$
and apply $\sin$ to both sides:
$$
x = \cos x \quad(1)
$$
Geometrically it means that you reflected the graph of these functions with respect to $y=x$. Now we can apply Contraction Mapping Theorem to solve $(1)$ on $[-1,1]$. You can show using Lipschitz continuity of $\cos$ that
$$
|\cos x' - \cos x''|\leq \alpha|x'-x''|
$$
where $\alpha = \sin1<1$. Now, we put $x_0 =0$ and construct $x_{n+1} = \cos x_n$ which converges to the solution $x^* = \lim\limits_{n\to\infty}x_n$. We only need to find bounds:
$$
|x^*-x_n| \leq \sum\limits_{k=n}^\infty|x_{k+1}-x_k| \leq \sum\limits_{k=n}^\infty\alpha^k = \frac{\alpha^n}{1-\alpha}
$$
where we used that $|x_{k+1}-x_k|\leq \alpha|x_k-x_{k-1}|\leq\dots\leq\alpha^k|x_1-x_0|=\alpha^k$. By the way, $\alpha\leq 0.85$
