Find the maximum and minimum value of `6sin x cos + 4 cos 2x`.

Answer

To find the maximum and minimum values of the expression \( 6 \sin x \cos x + 4 \cos 2x \), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression using the double angle identity for sine:
\[ 6 \sin x \cos x = 3 \cdot 2 \sin x \cos x = 3 \sin 2x \]
Thus, the expression becomes:
\[ 3 \sin 2x + 4 \cos 2x \]

### Step 2: Identify the coefficients
In the expression \( 3 \sin 2x + 4 \cos 2x \), we can identify the coefficients:
- \( a = 3 \)
- \( b = 4 \)

### Step 3: Use the formula for maximum and minimum values
The maximum and minimum values of the expression \( a \sin \theta + b \cos \theta \) can be found using the formula:
\[ \text{Maximum value} = \sqrt{a^2 + b^2} \]
\[ \text{Minimum value} = -\sqrt{a^2 + b^2} \]

### Step 4: Calculate \( a^2 + b^2 \)
Now, we calculate \( a^2 + b^2 \):
\[ a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

### Step 5: Find the maximum and minimum values
Now we can find the maximum and minimum values:
\[ \text{Maximum value} = \sqrt{25} = 5 \]
\[ \text{Minimum value} = -\sqrt{25} = -5 \]

### Conclusion
Thus, the maximum value of the expression \( 6 \sin x \cos x + 4 \cos 2x \) is \( 5 \) and the minimum value is \( -5 \).

---