Answer:
$0, \frac{\sqrt{5}-1}{2}$

Solution: Note that if $\sin (\theta)=0$, then the polynomial has 1 root. Now assume this is not the case then the polynomial is a quadratic in $x$.
Factor the polynomial as $(\tan (\theta) x+1)(x+\sec (\theta))$. Then the condition is equivalent to $\sec (\theta)=\frac{1}{\tan (\theta)}$, which is equivalent to $\sin (\theta)=\cos ^{2}(\theta)=1-\sin ^{2}(\theta)$. Solving now gives $\sin (\theta)=\frac{\sqrt{5}-1}{2}$ as the only solution.
16. [10] Given an angle $\theta$, consider the polynomial
$$
P(x)=\sin (\theta) x^{2}+(\cos (\theta)+\tan (\theta)) x+1 \text {. }
$$

Given that $P$ only has one real root, find all possible values of $\sin (\theta)$.