= Solution
For fixed $\mu\ne0$, a large positive root must lie close to a <pole> of the <tangent function>: away from its <poles>, $\tan x$ cannot balance the unbounded right side. Label the adjacent <poles> by $a_n=(n+1/2)\pi$ and write $x_n=a_n+\delta_n$. The <Taylor series> of $-\cot\delta$ gives
$$
-\frac1\delta+\frac\delta3+\frac{\delta^3}{45}+\cdots=\mu(a_n+\delta).
$$
Seek the <asymptotic expansion> $\delta=A/a_n+B/a_n^3+\cdots$. The coefficients of $a_n$ and $a_n^{-1}$ give
$$
-\frac1A=\mu,\qquad \frac B{A^2}+\frac A3=\mu A.
$$
Thus the <large roots near tangent poles> satisfy
$$
\boxed{x_n=a_n-\frac1{\mu a_n}+\frac{1/(3\mu^3)-1/\mu^2}{a_n^3}+O(a_n^{-5})}.
$$
The first correction already supplies the requested dependence on $\mu$. For $\mu>0$ the root approaches the <pole> from below; for $\mu<0$ it approaches from above. This labels roots by their nearby <poles>, avoiding an irrelevant finite shift in the enumeration of positive roots.
\b[No positivity restriction on $\mu$ is required.] The expansion requires $\mu\ne0$ fixed and $|\mu|a_n\gg1$, so the displacement is small. It is not uniform as $\mu\to0$. At $\mu=0$ the roots are exactly $n\pi$, a different leading sequence; these cannot be recovered by setting $\mu=0$ in the <pole> expansion. If $\mu$ varies with $n$, its size must be checked against the small-displacement and successive-term conditions rather than using the fixed-parameter remainder blindly.
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