<s>[INST] 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?
 [/INST] $ 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 $x=\dfrac{3 \pi}{2}$ 
</s>