Answer
$2\sin(x+\frac{5\pi}{6})$
Work Step by Step
Use the identity $A\sin x+B\cos x=k\sin(x+\phi)$, where $k=\sqrt{A^2+B^2}$ and $\phi$ satisfies $\cos \phi=\frac{A}{\sqrt{A^2+B^2}}$ and $\sin \phi=\frac{B}{\sqrt{A^2+B^2}}$.
In this case, $k=\sqrt{(-\sqrt{3})^2+1^2}=\sqrt{3+1}=\sqrt{4}=2$.
$\phi$ satisfies $\cos \phi=\frac{-\sqrt{3}}{\sqrt{(-3)^2+1^2}}=-\frac{\sqrt{3}}{2}$ and $\sin \phi=\frac{1}{\sqrt{(-3)^2+1^2}}=\frac{1}{2}$. The only values $\phi$ in $[0, 2\pi)$ where $\cos \phi=-\frac{\sqrt{3}}{2}$ are $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$, and of these two values, only $\phi=\frac{5\pi}{6}$ also satisfies $\sin\phi=\frac{1}{2}$, so $\phi=\frac{5\pi}{6}$.
Therefore, $-\sqrt{3}\sin x+\cos x=2\sin(x+\frac{5\pi}{6})$.
