# How do you find the points where the graph of the function  f(x)=sin2x+sin^2x has horizontal tangents?

Mar 15, 2016

Horizontal tangent means neither increasing nor decreasing. Specifically, the derivative of the function has to be zero $f ' \left(x\right) = 0$.

#### Explanation:

$f \left(x\right) = \sin \left(2 x\right) + {\sin}^{2} x$

$f ' \left(x\right) = \cos \left(2 x\right) \left(2 x\right) ' + 2 \sin x \cdot \left(\sin x\right) '$

$f ' \left(x\right) = 2 \cos \left(2 x\right) + 2 \sin x \cos x$

Set $f ' \left(x\right) = 0$

$0 = 2 \cos \left(2 x\right) + 2 \sin x \cos x$

$2 \sin x \cos x = - 2 \cos \left(2 x\right)$

$\sin \left(2 x\right) = - 2 \cos \left(2 x\right)$

$\sin \frac{2 x}{\cos} \left(2 x\right) = - 2$

$\tan \left(2 x\right) = - 2$

$2 x = \arctan \left(2\right)$

$x = \frac{\arctan \left(2\right)}{2}$

$x = 0.5536$

This is one point. Since solution was given out by $\tan$ , other points will be every π times the factor in $2 x$ meaning 2π. So the points will be:

x=0.5536+2n*π

Where $n$ is any integer.
graph{sin(2x)+(sinx)^2 [-10, 10, -5, 5]}