= Solution
Linearity and $\|Tf\|_\infty\leq2\pi\|f\|_\infty$ show that the <integral> defines a <bounded linear operator>. Inductively,
$$
\boxed{|T^kf(t)|\leq\frac{t^k}{k!}\|f\|_\infty\qquad(0\leq t\leq2\pi).}
$$
For $k=0$ this is immediate. If it holds for $k$, then $|\sin s|\leq1$ gives $|T^{k+1}f(t)|\leq\int_0^t s^k\|f\|_\infty/k!\,ds$, which proves the next case. Thus $\|T^k\|\leq(2\pi)^k/k!$.
For every $\lambda\ne0$, the <series> $A_\lambda=\sum_{k\geq0}\lambda^{-k-1}T^k$ converges in <operator norm>, dominated by $|\lambda|^{-1}e^{2\pi/|\lambda|}$. Multiplying its <partial sums> by $\lambda I-T$ on either side gives the identity minus a remainder $\lambda^{-N-1}T^{N+1}$, which tends to zero by the factorial estimate. Therefore $A_\lambda$ is the bounded inverse of $\lambda I-T$.
Finally every function in the <image> of $T$ vanishes at zero, so $T$ is not onto the <continuous functions>. Hence $0$ belongs to its <spectrum> and
$$
\boxed{\sigma(T)=\{0\}.}
$$
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