Answer
concave up
$\frac{\pi}{4}$ < $θ$ < $0$, $\frac{3\pi}{4}$ < $θ$ < $\pi$
concave down
$\frac{\pi}{4}$ < $θ$ < $\frac{3\pi}{4}$
inflection point at
$θ = \frac{\pi}{4}, \frac{3\pi}{4}$
Work Step by Step
Let
$f(θ)$ = $θ+sin^{2}θ$
$f'(θ)$ = $1+2sinθcosθ$ = $1+sin2θ$
$f''(θ)$ = $2cos2θ$
concave up
$2cos2θ$ > $0$
$cos2θ$ > $0$
$\frac{\pi}{2}$ < $2θ$ < $0$, $\frac{3\pi}{2}$ < $2θ$ < $2\pi$
$\frac{\pi}{4}$ < $θ$ < $0$, $\frac{3\pi}{4}$ < $θ$ < $\pi$
concave down
$\frac{\pi}{4}$ < $θ$ < $\frac{3\pi}{4}$
inflection point at
$θ = \frac{\pi}{4}, \frac{3\pi}{4}$