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.

If $f(x)=2x+5\cos x , x \in [0,2\pi], f'(x)=2-5 \sin x, f''(x)=-5 \cos x$
$$f'(x)=0 \implies \sin x=\frac{2}{5} \implies x_1= \sin^{-1} (2/5), x_2=\pi-\sin^{-1} (2/5)$$
$$f''(x_1)<0, f''(x_2)>0 \implies f_{max}=f(\sin^{-1}(2/5))=2\sin^{-1}(2/5)+\sqrt{21},$$ $$ f_{min}
=f(x_2)=2[\pi-\sin^{-1}(2/5)]-\sqrt{21}$$
$$f(0)=5, f[2\pi]=4\pi+5$$
So the absolute max is $f(2\pi)=4\pi+5$ and absolute min is $2[\pi-\sin^{-1}(2/5)]-\sqrt{21}$
