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.
 A: I'm posting this answer in response to the comment thread under picakhu's answer; writing comments was getting a bit tedious.  The answer has the same content as Isaac's, but is explained a little differently, which might be useful for the OP.
The general problem, of which this is a special case, is:
Given an expression of the form  $a \sin x + b \cos x$, for some numbers $a$ and $b$, to rewrite it in the form $c \sin (x + \theta)$, for some appropriate choice 
of $c$ and $\theta$ (which will be related to $a$ and $b$ in some way, of course --- and we have to find out what that way is!).
Suppose first that $a^2 + b^2 = 1$.  Then trigonometry (especially the point of view of the unit circle) tells us that there is an angle $\theta$ such
that $a = \cos \theta$ and $b = \sin \theta$.  
Thus we can write $a \sin x + b \cos x = \cos \theta \sin x + \sin \theta \cos x,$ and via the addition theorem for $\sin$, we recognize this as being $\sin(x + \theta).$
Now in most examples, it may not be that $a^2 + b^2 = 1$.  But it equals something (!), so let's call that something $c^2$.  Then we see
that $(a/c)^2 + (b/c)^2 = 1$, so we can choose $\theta$ so that
$a/c = \cos \theta$ and $b/c = \sin \theta$.  Then as above
we find that $(a/c)\sin x + (b/c) \cos x = \sin (x + \theta),$
and so (finishing finally) we have
$$a \sin x + b \cos x = c \sin (x + \theta).$$
In the OP's example, $a = \sqrt{3}$ and $b = 1$, so $c = 2$,
and we are led to the given answer.  
Note that in practice this whole process will be easiest when $a/c$ and $b/c$ are easily recognized trig function special values, like $1/2$ or $\sqrt{2}/2$ or $\sqrt{3}/2$.  If they are just somewhat random numbers, then you won't be able to figure out the correct choice of $\theta$ without using a calculator to
compute $\theta$.   
A: Instead of telling you how your teacher got that, try doing the opposite and expand out $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$
The rest should be trivial
A: 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:
$$\begin{align}
a\sin x+b\cos x
&=a\left(\sin x+\frac{b}{a}\cos x\right)
\\
&=a\left(\sin x+\tan\left(\arctan\frac{b}{a}\right)\cos x\right)
\\
&=a\left(\sin x+\frac{\sin\left(\arctan\frac{b}{a}\right)}{\cos\left(\arctan\frac{b}{a}\right)}\cos x\right)
\\
&=\frac{a}{\cos\left(\arctan\frac{b}{a}\right)}\left(\sin x\cos\left(\arctan\frac{b}{a}\right)+\sin\left(\arctan\frac{b}{a}\right)\cos x\right)
\\
&=\sqrt{a^2+b^2}\sin\left(x+\arctan\frac{b}{a}\right).
\end{align}$$
