finding a constant $c$ so that $ \hat{f} (m)=0 $ Let be $ c \in (0, 2 \pi) $
and 
$$f_c(x):= \begin{cases} \frac{x}{c} , 0 \leq x \leq c \\ \frac{2 \pi -x}{2 \pi -c}, c < x \leq 2 \pi \end{cases} $$
I want to determine $c$,
so that 
$\hat{f} (m)=0  $
for $ m \in 7 \mathbb{Z} \backslash \{ 0\} $
it is 
$ \hat{f}(m)= \frac{1}{2 \pi} \int_0^{2 \pi} f(x) e^{-imx} dx $
$= \frac{1}{2 \pi} [\int_0^c \frac{x}{c} e^{-imx} dx + \int_c^{2 \pi} \frac{2 \pi -x}{2 \pi -c} e^{-imx} dx] $
=$\frac{1}{2 \pi}[ [ \frac{(imx+1) e^{-imx}}{cm^2}  ]_0^c +[ \frac{(im(x- 1 \pi)+1)e^{-imx}}{(c-2 \pi )m^2}]_c^{2 \pi} $
=$ \frac{1}{2 \pi} [ \frac{(cm-i)sin(cm)+(icm+1)cos(cm)-1}{cm^2}] + \frac{1}{2 \pi} [- \frac{((c- 2 \pi)m-i)sin(cm)+((ic-2i \pi )m+1)cos(cm)+isin(2 \pi m)-cos(2 \pi m)}{(c-2\pi)m^2}] $
i was trying to reshape the equation $ \hat{f} (m) =! 0 $
but i get stuck at determining a $c$.
And what will it mean for $ m \in 7 \mathbb{Z} \backslash \{0 \} $
do you see a mistake here? Or is there an other approach?
Would be very thankful for any help!

We have
\begin{align}
\hat{f}(m) &= \frac{1}{2\pi}\int_0^c \frac{x}{c}\mathrm{e}^{-\mathrm{i}mx} dx
+ \frac{1}{2\pi}\int_c^{2\pi} \frac{2\pi - x}{2\pi - c}\mathrm{e}^{-\mathrm{i}mx} dx\\
&= \frac{1}{2\pi (2\pi - c) m^2}\left(1 - \mathrm{e}^{-\mathrm{i}2m\pi}\right)
- \frac{1}{(2\pi - c) cm^2}\left(1 - \mathrm{e}^{-\mathrm{i}cm}\right).
\end{align}
Remark: I used Maple to simplify the expressions.
For $m \in 7 \mathbb{Z} \backslash \{ 0\}$, we have
$$\hat{f}(m) = -\frac{1}{(2\pi - c) cm^2}\left(1 - \mathrm{e}^{-\mathrm{i}cm}\right).$$
Since $\hat{f}(m) = 0$ for all $m \in 7 \mathbb{Z} \backslash \{ 0\}$, 
we have $c = \frac{2\pi}{7}, \frac{4\pi}{7}, \frac{6\pi}{7}, \frac{8\pi}{7}, \frac{10\pi}{7}, \frac{12\pi}{7}$.
