HMMT February

We first claim that $\alpha:=1+\cos ^{2} x+\sin ^{4} x=1+\sin ^{2} x+\cos ^{4} x$. Indeed, note that $$\sin ^{4} x-\cos ^{4} x=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)=\sin ^{2} x-\cos ^{2} x$$ which is the desired after adding $1+\cos ^{2} x+\cos ^{4} x$ to both sides. Hence, since $\sin \alpha=\frac{13}{14}$, we have $\cos \alpha= \pm \frac{3 \sqrt{3}}{14}$. It remains to determine the sign. Note that $\alpha=t^{2}-t+2$ where $t=\sin ^{2} x$. We have that $t$ is between 0 and 1 . In this interval, the quantity $t^{2}-t+2$ is maximized at $t \in\{0,1\}$ and minimized at $t=1 / 2$, so $\alpha$ is between $7 / 4$ and 2 . In particular, $\alpha \in(\pi / 2,3 \pi / 2)$, so $\cos \alpha$ is negative. It follows that our final answer is $-\frac{3 \sqrt{3}}{14}$.