How does teacher get first step? Below are the steps the teacher took to solve:
$y = \sqrt{3}\sin x + \cos x$ find min and max on $[0, 2\pi)$
Step 1: = $2\sin(x + \pi/6))$
How does the teacher get this first step?
Note: No calculus please.

picakhu's answer is the simplest way to see how it works having already arrived at $y=2\sin(x+\frac{\pi}{6})$ (use the identity there to expand this form).  In general, given $a\sin x+b\cos x$ (let's say for $a,b>0$), it is possible to arrive at a similar equivalent form:
$$\begin{align}
a\sin x+b\cos x
&=a\left(\sin x+\frac{b}{a}\cos x\right)
\\
&=a\left(\sin x+\tan\left(\arctan\frac{b}{a}\right)\cos x\right)
\\
&=a\left(\sin x+\frac{\sin\left(\arctan\frac{b}{a}\right)}{\cos\left(\arctan\frac{b}{a}\right)}\cos x\right)
\\
&=\frac{a}{\cos\left(\arctan\frac{b}{a}\right)}\left(\sin x\cos\left(\arctan\frac{b}{a}\right)+\sin\left(\arctan\frac{b}{a}\right)\cos x\right)
\\
&=\sqrt{a^2+b^2}\sin\left(x+\arctan\frac{b}{a}\right).
\end{align}$$
