To find all points on the graph of the function $f(x) = 2\sin(x) + \sin^2(x)$ at which the tangent line is horizontal, we need to follow these steps:

Step 1: Find the derivative of the function

The tangent line is horizontal when the derivative of the function is zero. Let's differentiate $f(x) = 2\sin(x) + \sin^2(x)$ with respect to $x$.

$f'(x) = \frac{d}{dx}\left( 2\sin(x) \right) + \frac{d}{dx}\left( \sin^2(x) \right)$

Using the chain rule for $\sin^2(x)$, we get:

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

Simplifying:

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

Step 2: Set the derivative equal to zero

We now set the derivative equal to zero to find the points where the tangent line is horizontal:

$2\cos(x)(1 + \sin(x)) = 0$

This equation is satisfied if either factor is zero:

1. $\cos(x) = 0$
2. $1 + \sin(x) = 0$

Case 1: $\cos(x) = 0$

The cosine function equals zero at:

$x = \frac{\pi}{2} + n\pi \quad \text{for any integer } n$

Case 2: $1 + \sin(x) = 0$

Solving for $\sin(x)$, we get:

$\sin(x) = -1$

The sine function equals -1 at:

$x = \frac{3\pi}{2} + 2n\pi \quad \text{for any integer } n$

Step 3: Find the corresponding $y$-values

We now substitute the $x$-values back into the original function $f(x) = 2\sin(x) + \sin^2(x)$ to find the corresponding $y$-values.

For $x = \frac{\pi}{2} + n\pi$:

At $x = \frac{\pi}{2}$, $\sin\left(\frac{\pi}{2}\right) = 1$:

$f\left(\frac{\pi}{2}\right) = 2(1) + (1)^2 = 2 + 1 = 3$

At $x = \frac{3\pi}{2}$, $\sin\left(\frac{3\pi}{2}\right) = -1$:

$f\left(\frac{3\pi}{2}\right) = 2(-1) + (-1)^2 = -2 + 1 = -1$

Thus, the points where the tangent line is horizontal are:

(x, y) = \left(\frac{\pi}{2} + n\pi, 3\right) \quad \text{(larger y-value)} (x, y) = \left(\frac{3\pi}{2} + 2n\pi, -1\right) \quad \text{(smaller y-value)}

Final Answer:

1. $(x, y) = \left(\frac{\pi}{2}, 3\right)$ (smaller $y$-value)
2. $(x, y) = \left(\frac{3\pi}{2}, -1\right)$ (larger $y$-value)

Let me know if you would like to discuss the steps in more detail.

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Here are 5 related questions to explore:

1. How would the process change if we had $f(x) = \sin(x) + \sin^2(x)$ instead?
2. What is the geometric interpretation of a horizontal tangent line?
3. Can you find points where the tangent line has a slope of 1?
4. How would the second derivative help in confirming the type of extremum?
5. What if we had $f(x) = 2\sin(x) + \cos^2(x)$? Would the solution be similar?

Tip: Always check critical points by plugging them back into the original equation to ensure correctness.