$\begin{array}{1 1}\frac{\tan \theta}{k}\\k\;tan\;\theta\\\frac{\cot \theta}{k} \\k\;cot \theta\end{array} $

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Answer : $\large\frac{\tan \theta}{k}$

The net downward force on the body at a distance x is

$f(x) =mg \sin \theta - \mu mg \cos \theta$

$\qquad= mg( \sin \theta - \mu \cos \theta)$

$\qquad= mg (\sin \theta- kx \cos \theta)$

$\therefore f(x) =0$ at a value of $x =x_0$ given by

$\sin \theta -kx_0 \cos \theta=0$

Which gives $x_0 = \large\frac{\tan \theta}{k}$

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