HMMT-Nov 2022 guts, Problem 16

Given an angle $\theta$, consider the polynomial \[ P(x)=\sin (\theta) x^{2}+(\cos (\theta)+\tan (\theta)) x+1 \] Given that $P$ only has one real root, find all possible values of $\sin (\theta)$.

Solution: 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. $\fbox{0, \frac{\sqrt{5}-1}{2}}$.

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