Suppose  x  is a real number such that  $\text{sin}(1+{\text{cos}}^{2}x+{\text{sin}}^{4}x)=\frac{13}{14}$. Compute $\text{cos}(1+{\text{sin}}^{2}x+{\text{cos}}^{4}x)=\frac{a\sqrt{3}}{b}\text{\hspace{0.17em}}where\left|a\right|\&\left|b\right|$  are relatively prime number then the value of  $b-a$ equals

We first claim that  $\alpha :=1+{\text{cos}}^{2}x+{\text{sin}}^{4}x=1+{\text{sin}}^{2}x+{\text{cos}}^{4}x$. Indeed, note that ${\text{sin}}^{4}x-{\text{cos}}^{4}x=({\text{sin}}^{2}x+{\text{cos}}^{2}x)({\text{sin}}^{2}x-{\text{cos}}^{2}x)={\text{sin}}^{2}x-{\text{cos}}^{2}x$
which is the desired after adding $1+{\text{cos}}^{2}x+{\text{cos}}^{4}x$  to both sides.
Hence,  $\text{sincesin}\alpha =\frac{13}{14}$, we have  $\text{cos}\alpha =\pm \frac{3\sqrt{3}}{14}$. It remains to determine the sign. Note that  $\alpha =t^2-t+2$ where  $t={\text{sin}}^{2}x$. 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 $-\frac{3\sqrt{3}}{14}$.