Define the function $x \in L^2[-\pi, \pi]$ by \begin{equation*} x(t) = \left\{ \begin{aligned} 0 , && -\pi \leq t < 0, \\ 1 , && 0\leq t \leq \pi \end{aligned} \right. \end{equation*} and the operator $T : L^2[-\pi, \pi] \to L^2[-\pi, \pi]$ by setting
\begin{equation} Tf = \langle f, e_0\rangle e_0 + 3\langle f, e_{-1}\rangle e_{-1} + 3\langle f, e_1\rangle e_1\,, \end{equation}
where $e_n(t) = \frac 1 {\sqrt{2\pi}} \mathrm e^{\mathrm i n t}$ for $n \in \mathbb Z$.
Show that the integers $0$, $1$ and $3$ are the only eigenvalues of $T$.
An answer that came up short
In the (a) part of this question, I came up with the basis representation of $x \in L^2$:
\begin{equation} x = \sqrt{ \frac{ \pi }{ 2 } } + \frac 1 { \sqrt{2 \pi} }\sum_{n \neq 0} \left( \frac{ \mathrm e^{\mathrm i n \pi } - 1 }{ \mathrm i n } \right) e_n \,. \end{equation}
I then proceeded to find out what $Tf$ looks like: \begin{align*} Tf &= \langle f, e_0\rangle e_0 + 3 \langle f, e_{-1}\rangle e_{-1} + 3\langle f, e_1\rangle e_1 \\ &= \sqrt{\frac{ \pi }{ 2 } } \frac{ 1 }{ \sqrt{ 2\pi } } + \frac 3 { \sqrt{2 \pi} } \left( \frac{ \mathrm e^{-\mathrm i \pi } - 1 }{ - \mathrm i } \right) \frac 1 {\sqrt{2\pi}} \mathrm e^{- \mathrm i t} + \frac 3 { \sqrt{2 \pi} } \left( \frac{ \mathrm e^{ \mathrm i \pi} - 1 }{ \mathrm i } \right) \frac 1 {\sqrt{2\pi}} \mathrm e^{ \mathrm i t} \\ &= \frac 1 2 + \frac 3 { 2 \pi } \left( \frac{ -2 }{ - \mathrm i } \right) \mathrm e^{- \mathrm i t} + \frac 3 { 2 \pi } \left( \frac{ -2 }{ \mathrm i } \right) \mathrm e^{ \mathrm i t} = \frac 1 2 + \frac {6} { 2\pi\mathrm i } \mathrm e^{- \mathrm i t} - \frac {6 } {2\pi\mathrm i } \mathrm e^{\mathrm i t}\\ &= \frac 1 2 + \frac {6} { \pi } \frac{ \mathrm e^{- \mathrm i t} - \mathrm e^{ \mathrm i t} }{ 2\mathrm i } = \frac 1 2 - \frac {6} { \pi } \frac{ \mathrm e^{ \mathrm i t} - \mathrm e^{- \mathrm i t} }{ 2\mathrm i } = \frac 1 2 - \frac {6} { \pi } \sin{t} \\ &= \left\{ \begin{aligned} 0 , && -\pi \leq t < 0\,, \\ 1 , && 0\leq t \leq \pi \,. \end{aligned} \right. \end{align*}
In other words, \begin{equation} \sin{t} = \frac \pi {12} \end{equation} when $t \in [-\pi,0)$ and when $t \in [0, \pi]$ \begin{equation} \sin{t} = -\frac{ \pi }{ 12 }, \end{equation} which is not possible as $\sin t$ is not constant on either of these two intervals.
Why is this approach not working?