Show that $f(x)\equiv 0$. 
Let $f:[0,2\pi]\to\mathbb{R}$, which is $2\pi$ periodic and continuous. It is given that for every $n\in\mathbb{Z}$:$$\int_0^{2\pi} f(x)e^{i\left(n+\frac{1}{2}\right)x} = 0.$$ Show that $f(x)\equiv 0$.

So this expression is almost the $n$-th Fourier coefficient $$\hat f(n) = \frac{1}{2\pi} \int_0^{2\pi} f(x)e^{-inx}\ dx$$
I am almost certain there needs to be some substitution in order to express the above with $\hat f(n)$.
I'd be glad for help. 
 A: Consider $g(x) = f(x) \cdot e^{ix/2}$ and find out what the assumption tells you about $g$. Then find out what this tells you about $f$.
A: Hint: if $f(x)$ is $2\pi$-periodic, then $f(x) e^{ix/2}$ is $(\ldots)$-periodic. Fill the dots and use Fourier decomposition with such a periodicity. 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{f}\pars{x} & = \sum_{n = -\infty}^{\infty}a_{n}\expo{\ic nx}\iff
a_{n} = \int_{0}^{2\pi}\mrm{f}\pars{x}\expo{-\ic nx}\,{\dd x \over 2\pi}
\\[5mm] 
\mbox{Moreover,}\quad
\mrm{f}\pars{x} & =
\sum_{n = -\infty}^{\infty}\bracks{\int_{0}^{2\pi}\mrm{f}\pars{x'}
\expo{-\ic nx'}\,{\dd x' \over 2\pi}}\expo{\ic nx}
\\[5mm] &=
\int_{0}^{2\pi}\mrm{f}\pars{x'}\
\underbrace{\bracks{{1 \over 2\pi}\sum_{n = -\infty}^{\infty}
\expo{\ic n\pars{x - x'}}}}_{\ds{\delta\pars{x - x'}}}\ \dd x'
\end{align}

Therefore,

\begin{align}
0 & =
\int_{0}^{2\pi}\mrm{f}\pars{x}\expo{\ic\pars{n + 1/2}x}\,\dd x \implies
0 =
{1 \over 2\pi}\sum_{n = -\infty}^{\infty}\expo{-\ic ny}
\int_{0}^{2\pi}\mrm{f}\pars{x}\expo{\ic\pars{n + 1/2}x}\,\dd x
\\[5mm] \implies 0 & =
\int_{0}^{2\pi}\mrm{f}\pars{x}\expo{\ic x/2}\bracks{{1 \over 2\pi}
\sum_{n = -\infty}^{\infty}\expo{\ic n\pars{x - y}}}\dd x =
\int_{0}^{2\pi}\mrm{f}\pars{x}\expo{\ic x/2}\delta\pars{x - y}\,\dd x
\\[5mm] & =
\mrm{f}\pars{y}\expo{\ic y/2} \implies \bbx{\ds{\mrm{f}\pars{y} = 0}}
\end{align}
