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To find the points where the tangent to the curve \( y = \cos(2x) + 2\cos(x) \) is horizontal, we need to determine where the derivative of \( y \) with respect to \( x \), \( \frac{dy}{dx} \), is equal to zero. This is because a horizontal tangent implies that the slope of the tangent line is zero.

### Step-by-Step Solution

1. **Find \( \frac{dy}{dx} \):**
   Given:
   \[
   y = \cos(2x) + 2\cos(x)
   \]
   Differentiate \( y \) with respect to \( x \):
   - The derivative of \( \cos(2x) \) is \( -2\sin(2x) \) (using the chain rule).
   - The derivative of \( 2\cos(x) \) is \( -2\sin(x) \).

   So:
   \[
   \frac{dy}{dx} = -2\sin(2x) - 2\sin(x)
   \]

2. **Simplify \( \frac{dy}{dx} \):**
   Using the double-angle identity \( \sin(2x) = 2\sin(x)\cos(x) \), we can rewrite \( -2\sin(2x) \) as:
   \[
   -2(2\sin(x)\cos(x)) = -4\sin(x)\cos(x)
   \]
   Thus:
   \[
   \frac{dy}{dx} = -4\sin(x)\cos(x) - 2\sin(x)
   \]
   Factor out \( -2\sin(x) \):
   \[
   \frac{dy}{dx} = -2\sin(x)(2\cos(x) + 1)
   \]

3. **Set \( \frac{dy}{dx} = 0 \):**
   For the tangent to be horizontal, we set \( \frac{dy}{dx} = 0 \):
   \[
   -2\sin(x)(2\cos(x) + 1) = 0
   \]
   This equation is satisfied if either \( \sin(x) = 0 \) or \( 2\cos(x) + 1 = 0 \).

4. **Solve for \( x \):**

   - **Case 1:** \( \sin(x) = 0 \)
     - \( \sin(x) = 0 \) when \( x = 0 \) or \( x = \pi \) (within the interval \( [0, 2\pi] \)).

   - **Case 2:** \( 2\cos(x) + 1 = 0 \)
     - Solving for \( \cos(x) \):
       \[
       \cos(x) = -\frac{1}{2}
       \]
     - \( \cos(x) = -\frac{1}{2} \) when \( x = \frac{2\pi}{3} \) or \( x = \frac{4\pi}{3} \) (within the interval \( [0, 2\pi] \)).

5. **Conclusion:**
   The points at which the tangent is horizontal on the interval \( [0, 2\pi] \) are:
   \[
   x = 0, \; x = \pi, \; x = \frac{2\pi}{3}, \; x = \frac{4\pi}{3}
   \]

Would you like further details on any of these steps or have any additional questions?

### Related Questions

1. How would the result change if we altered the function to \( y = \sin(2x) + 2\sin(x) \)?
2. What are the key properties of functions that allow horizontal tangents to occur?
3. How does the double-angle identity assist in simplifying trigonometric derivatives?
4. Can you explain how to determine the interval of horizontal tangents for different trigonometric functions?
5. How would the answer change if the interval were extended beyond \( [0, 2\pi] \)?

### Tip
Always check the interval when solving for horizontal tangents, as solutions outside the specified range may need to be excluded.

Find the points (in exact form) at which the tangent is horizontal on the interval y = cos(2x) + 2cos(x), [0, 2π]

Discover the points at which the tangent to the curve y = cos(2x) + 2cos(x) is horizontal on the interval [0, 2π]. This problem involves taking derivatives, applying trigonometric identities, and solving for x to find the exact points where the slope is zero. Suitable for Grades 11-12.

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