= Solution
Since
$$
\mathcal F^{-1}[e^{ika}]=\delta(x+a),
\qquad
\mathcal F^{-1}[e^{-ika}]=\delta(x-a),
$$
the <Inverse Fourier transforms of phase factors> give
$$
\boxed{
\mathcal F^{-1}[\cos(ka)]
=\frac12\bigl[\delta(x+a)+\delta(x-a)\bigr]
}
$$
and
$$
\boxed{
\mathcal F^{-1}[\sin(ka)]
=\frac1{2i}\bigl[\delta(x+a)-\delta(x-a)\bigr].
}
$$
From $z=\tan w=-i(e^{2iw}-1)/(e^{2iw}+1)$ one obtains
$$
e^{2iw}=\frac{1+iz}{1-iz},
\qquad
w=\frac1{2i}\log\frac{1+iz}{1-iz}.
$$
For $z=(2\sqrt3-3i)/7$ the quotient is $1+i\sqrt3=2e^{i\pi/3}$. Its principal logarithm is $\log2+i\pi/3$, so the principal value is
$$
\boxed{\tan^{-1}z=\frac\pi6-\frac i2\log2.}
$$
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