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.