Determine the coordinates of any max/min points on functions between 0 ≤ x ≤ 2π Ok so i'm working with trig functions on graphs and how to work out the amplitude, period and maximum/minimum turning points etc. 
One of the equations is: y = sin(2x) + 3
I need to find the coordinates of any max/min turning points for the function between 0 ≤ x ≤ 2π (so the answer is in radians).
In the book the answer is:
max at ((π/4),4) and at ((5π/4),4)
min at ((3π/4),2) and at ((7π/4),2)
Could someone please explain in a simple way how to achieve this result and how to solve questions similarly. 
Thank you.  
 A: Use stretching and shift properties of $\sin x$. You can find $x$ co-ordinates for the maxima/minima like this, then substitute in to give the solution. Below is a spoiler on how to do it for this question.

 You should know that $\sin x$ has maxima at $x=\pi/2,5\pi/2$ and minima at $x=3\pi/2,7\pi/2$. So $\sin2x$ would have maxima at $x=\pi/4,5\pi/4$ and minima at $x=3\pi/4,7\pi/4$. Now note that $y=\sin2x+3$ is just the same as $y=\sin 2x$, but shifted upwards in the $y$ direction, so the x co-ordinates of the maxima and minima are the same for both graphs. Finally, substituting in the values of $x$ which give maxima and minima will give you the $y$ co-ordinates, which are of course 4 and 2, as given in the question.

A: First you want to find where these maximums or minimums (called extrema in general) will occur, i.e. their x coordinates. Notice that normally, $\sin x$ has maximums when $x=\frac{\pi}{2}$ and when $x=\frac{3\pi}{2}$ because that's the top and bottom of the unit circle, where $\sin x$ has the largest magnitude. Now in this case, we want to find where $\sin(2x)+3$ has maximums. That $3$ (and any constants) will affect what the maximum is but not where it occurs. So it is sufficient to find where $\sin 2x$ has extrema. As mentioned previously, this will be where the inside of the sine function is $\frac{\pi}{2}$ or $\frac{3\pi}{2}$. So now we just solve:
$$2x=\frac{\pi}{2} \implies x=\frac{\pi}{4}$$
$$2x=\frac{3\pi}{2} \implies x=\frac{3\pi}{4}$$
But we know we want all solutions between $0$ and $2\pi$, which may include $2x>2\pi$, so we keep searching:
$$2x=\frac{5\pi}{2} \implies x=\frac{5\pi}{4}$$
$$2x=\frac{7\pi}{2} \implies x=\frac{7\pi}{4}$$
$$2x=\frac{9\pi}{2} \implies x=\frac{9\pi}{4}>2\pi$$
That last one is the one we don't care about, so we've found where all the extrema are in between $0$ and $2\pi$: $x=\frac{\pi}{4},\frac{3\pi}{4},\frac{4\pi}{4},\frac{7\pi}{4}$. Now we just find the actual values of these extrema by plugging those $x$ values in. After plugging and chugging we get the points $(\frac{\pi}{4},4),(\frac{3\pi}{4},2),(\frac{5\pi}{4},4),(\frac{7\pi}{4},2)$. Just looking at the $y$ values tells you which are maxima and which are minima.
