How do you determine the local extrema points for $y=\sqrt{3}\cos(3x)+\sin(3x)$ $$y=\sqrt{3}\cos(3x)+\sin(3x); 0\le{x}\le{\frac{2\pi}{3}}$$ 
I know that the local extrema can be determined by using the first derivative test. I took the derivative of $y$ and got $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x)$$ I then solved the derivative for when it's value is $0$ and got $x=\dfrac{\pi}{2}$ . I then used this critical point and subdivided the interval. I found that there were 3 points of local extrema after doing all my work, local maximum $x=0, \dfrac{2\pi}{3}$ and local minimum $x=\dfrac{\pi}{2}$. However according to the online homework, there were 4 different points of extrema. Local maximum $x=\dfrac{\pi}{18},\dfrac{2\pi}{3}$ and local minimum $x=0,\dfrac{7\pi}{18}$. I am really confused as to how there are 4 points of local extrema, did I leave out an answer somewhere? I am also confused as to how they got $x=\dfrac{\pi}{18},\dfrac{7\pi}{18}$ as points of local extrema. Could someone explain this to me?
 A: As I noted in my comment: At $x = \pi/2, \quad y' \neq 0;$ ... $y'(\pi/2) = 3\sqrt{3}$
You need to solve for $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x) = 0 \quad \iff \quad 3\sqrt{3}\sin(3x) = 3\cos(3x)$$
$$\iff \quad  \sqrt 3 \sin(3x) = \cos(3x),\quad x\in \left[0, \frac{2\pi}{3}\right]$$
Note that $$\sqrt 3 \sin(3x) = \cos(3x) \iff \sqrt 3 \dfrac{\sin(3x)}{\cos(3x)} = \sqrt 3 \tan(3x) = 1\iff \tan(3x) = \frac{1}{\sqrt 3}$$
Solving for $x$ will give you 2 potential critical points on your interval; then recall that you need to also check endpoints of an interval as potential extrema.
A: $ y=\sqrt{3}\cos(3x)+\sin(3x)=2(\dfrac{\sqrt{3}}{2}\cos(3x)+\dfrac{1}{2}\sin(3x)=2(\sin\dfrac{\pi}{3}\cos(3x)+cos\dfrac{\pi}{3}\sin(3x))=2\sin(3x+\dfrac{\pi}{3}) $
$\dfrac{\pi}{3} \leq 3x+\dfrac{\pi}{3} \leq 2\pi+\dfrac{\pi}{3}$, so there is 2 peaks when  $3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}$ or $\dfrac{3 \pi}{2} $, another 2 are the end points ie:$x=0$ or $x=\dfrac{3 \pi}{2}$ 
A: $y=\sqrt 3\cos(3x)+\sin(3x)=2\sin(3x+\frac\pi3)$
$y'=2\cdot3\cos(3x+\frac\pi3)$
For the extreme values of $y, y'=0\implies \cos(3x+\frac\pi3)=0$
$\implies  3x+\frac\pi3=(2n+1)\frac\pi2$ whether $n$ is any integer
$\implies 3x =(6n+1)\frac\pi6 $
As $0\le x\le\frac{2\pi}3, 0\le 3x\le 2\pi$
$\implies 0\le (6n+1)\frac\pi6\le 2\pi \implies0\le 6n+1\le 12\implies n=0,1$
