Find the maximum and minimum values of the function `f(x) = sin x + cos 2x`.

Answer

To find the maximum and minimum values of the function \( f(x) = \sin x + \cos 2x \), we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos 2x) \]

Using the derivatives of sine and cosine, we have:

\[ f'(x) = \cos x - 2\sin 2x \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:

\[ \cos x - 2\sin 2x = 0 \]

Using the double angle identity \( \sin 2x = 2\sin x \cos x \), we can rewrite the equation:

\[ \cos x - 4\sin x \cos x = 0 \]

Factoring out \( \cos x \):

\[ \cos x (1 - 4\sin x) = 0 \]

This gives us two cases to consider:

1. \( \cos x = 0 \)
2. \( 1 - 4\sin x = 0 \)

### Step 3: Solve for critical points
**Case 1:** \( \cos x = 0 \)

This occurs at:

\[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

**Case 2:** \( 1 - 4\sin x = 0 \)

Solving for \( \sin x \):

\[ \sin x = \frac{1}{4} \]

### Step 4: Evaluate the function at critical points
Now we will evaluate \( f(x) \) at the critical points found.

1. **For \( x = \frac{\pi}{2} \):**

\[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos\left(2 \cdot \frac{\pi}{2}\right) = 1 + \cos(\pi) = 1 - 1 = 0 \]

2. **For \( \sin x = \frac{1}{4} \):**

We find \( x \) using the arcsine function:

\[ x = \arcsin\left(\frac{1}{4}\right) + 2n\pi \quad \text{or} \quad x = \pi - \arcsin\left(\frac{1}{4}\right) + 2n\pi \]

Now we substitute \( \sin x = \frac{1}{4} \) into \( f(x) \):

Using the identity \( \cos 2x = 1 - 2\sin^2 x \):

\[ \cos 2x = 1 - 2\left(\frac{1}{4}\right)^2 = 1 - 2\left(\frac{1}{16}\right) = 1 - \frac{1}{8} = \frac{7}{8} \]

Thus,

\[ f(x) = \frac{1}{4} + \frac{7}{8} = \frac{2}{8} + \frac{7}{8} = \frac{9}{8} \]

### Step 5: Determine maximum and minimum values
From the evaluations:

- At \( x = \frac{\pi}{2} \), \( f\left(\frac{\pi}{2}\right) = 0 \)
- At \( x = \arcsin\left(\frac{1}{4}\right) \) or \( x = \pi - \arcsin\left(\frac{1}{4}\right) \), \( f(x) = \frac{9}{8} \)

Thus, the maximum value of \( f(x) \) is \( \frac{9}{8} \) and the minimum value is \( 0 \).

### Final Answer:
- Maximum value: \( \frac{9}{8} \)
- Minimum value: \( 0 \)