Answer
$F$ is the smallest when $\theta = tan^{-1}~\mu$
Work Step by Step
$F = \frac{\mu~W}{\mu~sin~\theta+cos~\theta}$
Note that $0 \leq \theta \leq \frac{\pi}{2}$
We can find the value of $\theta$ when $F'(\theta) = 0$:
$F(\theta) = \frac{\mu~W}{\mu~sin~\theta+cos~\theta}$
$F'(\theta) = \frac{-(\mu~W)(\mu~cos~\theta-sin~\theta)}{(\mu~sin~\theta+cos~\theta)^2} = 0$
$\mu~W~sin~\theta = \mu^2~W~cos~\theta$
$sin~\theta = \mu~cos~\theta$
$tan~\theta = \mu$
$\theta = tan^{-1}~\mu$
When $tan~\theta = \mu$:
$sin~\theta = \frac{\mu}{\sqrt{\mu^2+1}}$
$cos~\theta = \frac{1}{\sqrt{\mu^2+1}}$
We can find the value of $F$ when $tan~\theta = \mu$:
$F =\frac{\mu~W}{\mu~sin~\theta+cos~\theta}$
$F =\frac{\mu~W}{\mu\cdot\frac{\mu}{\sqrt{\mu^2+1}}+\frac{1}{\sqrt{\mu^2+1}}}$
$F =\frac{\mu~W~\sqrt{\mu^2+1}}{\mu^2+1}$
$F =\frac{\mu~W}{\sqrt{\mu^2+1}}$
We can verify the value of $F$ at the endpoints of the interval $[0, \frac{\pi}{2}]$:
$F(0) =\frac{\mu~W}{\mu~sin~0+cos~0} = \mu~W$
$F(\frac{\pi}{2}) =\frac{\mu~W}{\mu~sin~(\pi/2)+cos~(\pi/2)} = W$
Note that when $tan~\theta = \mu$:
$F =\frac{\mu~W}{\sqrt{\mu^2+1}} \lt \frac{\mu~W}{\sqrt{1}} = \mu~W$
$F =\frac{\mu~W}{\sqrt{\mu^2+1}} \lt \frac{\mu~W}{\sqrt{\mu^2}} = W$
Therefore, $F$ is the smallest when $\theta = tan^{-1}~\mu$
