MHB What values of x make the graph of f(x) have a horizontal tangent?

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The graph of the function f(x) = x + 2sin(x) has a horizontal tangent when its derivative f'(x) = 1 + 2cos(x) equals zero. Setting the derivative to zero leads to the equation 1 + 2cos(x) = 0, which simplifies to cos(x) = -1/2. The solutions for x occur in quadrants II and III, yielding the general solutions x = (2k+1)π ± π/3, where k is an integer. This indicates multiple values of x that satisfy the condition for a horizontal tangent. The discussion emphasizes the importance of understanding the cosine function and its periodic nature in finding these solutions.
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For what values of does the graph of have a horizontal
tangent?
f(x) = x + 2sin(x)

I get this:

$f'(x) = 1 + 2 \cos(x)$

And I understand that I need to set this to zero.

$1 + 2cosx = 0$

$cosx = 1/2$

How do I isolate x in this situation?
 
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First of all, the equations is cos(x) = -1/2;

arccos gives you cos(\pi/3) = 0.5;

If you draw the graph of cos(x), this might help find you all the values for x where cos(x) = -0.5
 
Yes, as mentioned, you wind up with:

$$\cos(x)=-\frac{1}{2}$$

Now, there is a quadrant II and a quadrant III solution, given generally by:

$$x=(2k+1)\pi\pm\frac{\pi}{3}=\frac{\pi}{3}\left(3(2k+1)\pm1\right)$$ where $$k\in\mathbb{Z}$$
 

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