To find the points on the graph of the function $f(x) = 2\sin(x) + \sin^2(x)$ where the tangent line is horizontal, we need to determine where the derivative of the function $f'(x)$ is equal to zero. A horizontal tangent line occurs where the slope is zero, which means the first derivative must equal zero.

Step 1: Differentiate the function

The given function is: $f(x) = 2\sin(x) + \sin^2(x)$ We will differentiate this with respect to $x$.

The derivative of $2\sin(x)$ is: $\frac{d}{dx}[2\sin(x)] = 2\cos(x)$

The derivative of $\sin^2(x)$ (using the chain rule) is: $\frac{d}{dx}[\sin^2(x)] = 2\sin(x)\cos(x)$

Thus, the derivative of $f(x)$ is: $f'(x) = 2\cos(x) + 2\sin(x)\cos(x)$

Step 2: Set the derivative equal to zero

We need to find where the tangent is horizontal, i.e., where $f'(x) = 0$: $2\cos(x) + 2\sin(x)\cos(x) = 0$

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

This equation will be zero when either factor is zero.

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

$\cos(x) = 0 \implies x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$

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

$1 + \sin(x) = 0 \implies \sin(x) = -1$ This occurs at: $x = \frac{3\pi}{2} + 2n\pi, \quad n \in \mathbb{Z}$

Step 3: Find corresponding $y$-values

Now that we have the $x$-coordinates, we find the corresponding $y$-values by substituting these values into the original function $f(x)$.

Case 1: $x = \frac{\pi}{2} + n\pi$

For $x = \frac{\pi}{2}$: $f\left(\frac{\pi}{2}\right) = 2\sin\left(\frac{\pi}{2}\right) + \sin^2\left(\frac{\pi}{2}\right) = 2(1) + 1^2 = 3$ Thus, the point is $\left(\frac{\pi}{2}, 3\right)$.

For $x = \frac{3\pi}{2}$: $f\left(\frac{3\pi}{2}\right) = 2\sin\left(\frac{3\pi}{2}\right) + \sin^2\left(\frac{3\pi}{2}\right) = 2(-1) + (-1)^2 = -2 + 1 = -1$ Thus, the point is $\left(\frac{3\pi}{2}, -1\right)$.

Final Answer:

The points at which the tangent line is horizontal are: $(x, y) = \left(\frac{\pi}{2}, 3\right)$ and $(x, y) = \left(\frac{3\pi}{2}, -1\right)$

Would you like more details or have any questions? Here are 5 related questions you might explore:

1. What is the general formula for finding horizontal tangents in a function?
2. How would the procedure change if the function involved $\cos^2(x)$ instead of $\sin^2(x)$?
3. What other types of critical points, besides horizontal tangents, can be found using derivatives?
4. How would the result change if the original function was $f(x) = 2\sin(x) + \cos^2(x)$?
5. How does the graph of $f(x) = 2\sin(x) + \sin^2(x)$ compare to simpler sine and cosine functions?

Tip: Always remember to check both the first and second derivative to classify the type of critical points!