<s>[INST] Finding the absolute extrema of $F(x) = 2x + 5\cos(x)$ 
Find the absolute extrema of $F(x) = 2x + 5\cos(x)$ on the interval $[0,2\pi]$ using the extreme value theorem.
Answer should be 2 ordered pairs.

I got $\arcsin(2/5)$ for the first value of $x$, but can’t figure out the second.
Thanks in advance.
 [/INST] The question says absolute extreme.
$5\cos x \le 5$ and $2x \le 4\pi$ so $2x+ 5\cos x\le 5+4\pi$ so if $2x + 5\cos x$ ever equals $5+4\pi$, which it does at $x = 2\pi$, that will be an absolute (albeit not necessarily a local) maximum.  So the absolute maximum is $(2\pi, 5+4\pi)$.
We can't do the same for minimum as $2x \ge 0$ and $5\cos x \ge -5$ but when $2x=0$ we do not have $5\cos x =-5$ for $x =0$ so $2x + 5\cos x > -5$.
For a minimum we can find local minimum via $[2x+5\cos x]'= 2 -5\sin x = 0$ so $\sin x =\frac 25$.  So $0<\frac 25 < 1$ there will be one such value $x_1$ in the first quadrant and a second value $x_2$ in the second quadrant.  $x_1$ is $\arcsin \frac 25$ and $x_2$ is $\pi-\arcsin \frac 25$.
As $\sin x_1 =\frac 25$ then $\cos x_1 = \sqrt{1 - \frac 4{25}} = \frac {\sqrt{21}}5$.  And $x_2$ being in the second quadrant has $\cos x_2 = -\frac {\sqrt {21}}5$.
So the local extrema are $(\arcsin \frac 25, 2\arcsin \frac 25 + \sqrt {21})$ and $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$
Now $\frac 25 < \frac 12$ so $0< \arcsin \frac 25 < \frac \pi 6$  and $5>\sqrt{21} > 4$ so $2\pi -2\arcsin \frac 25-\sqrt{21} < 2\pi - 4 < 7- 4 =3< \sqrt{21}<2\arcsin \frac 25 + \sqrt {21}< 4\pi + 5$
So $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$ is a local minimum and $(\arcsin \frac 25, 2\arcsin \frac 25 + \sqrt {21})$ is a local but NOT an absolute maximum.
Now $2*0 + 5\cos 0  =5> 3 >2\pi -2\arcsin \frac 25-\sqrt{21}$.  So the local minimum is the absolute minimum.
So the absolute extrema at the local minimum, $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$ and at the end point, $(2\pi, 4\pi + 5)$.
</s>