Answer
The points on the graph at which the tangent line is horizontal are:
$(\frac{\pi}{2}+2\pi~n, 3)~~~$ where $n$ is an integer
$(\frac{3\pi}{2}+2\pi~n, -1)~~~$ where $n$ is an integer
Work Step by Step
$f(x) = 2~sin~x+sin^2~x$
$f'(x) = 2~cos~x+2~sin~x~cos~x$
If the tangent line is horizontal at a point $x$, then the slope of the graph at $x$ is 0.
We can find the values of $x$ where $f'(x) = 0$:
$f'(x) = 2~cos~x+2~sin~x~cos~x = 0$
$2~cos~x(1+sin~x) = 0$
$cos~x=0~~~$ or $~~~1+sin~x = 0$
$x=\frac{\pi}{2}+\pi~n~~~$ where $n$ is an integer
or
$x=\frac{3\pi}{2}+\pi~n~~~$ where $n$ is an integer
When $x = \frac{\pi}{2}$:
$f(\frac{\pi}{2}) = 2~sin~\frac{\pi}{2}+sin^2~\frac{\pi}{2}$
$f(\frac{\pi}{2}) = 2+1$
$f(\frac{\pi}{2}) = 3$
When $x = \frac{3\pi}{2}$:
$f(\frac{3\pi}{2}) = 2~sin~\frac{3\pi}{2}+sin^2~\frac{3\pi}{2}$
$f(\frac{3\pi}{2}) = -2+1$
$f(\frac{3\pi}{2}) = -1$
The points on the graph at which the tangent line is horizontal are:
$(\frac{\pi}{2}+2\pi~n, 3)~~~$ where $n$ is an integer
$(\frac{3\pi}{2}+2\pi~n, -1)~~~$ where $n$ is an integer
