Solve `sin 2x+cos 4x=2`.

Explanation

To solve the equation \( \sin 2x + \cos 4x = 2 \), we can follow these steps:

### Step 1: Analyze the equation
The equation is given as:
\[ \sin 2x + \cos 4x = 2 \]
We know that the maximum value of \( \sin \) and \( \cos \) functions is 1. Therefore, the maximum value of \( \sin 2x + \cos 4x \) can be at most \( 1 + 1 = 2 \).

### Step 2: Determine when the maximum value occurs
For \( \sin 2x + \cos 4x \) to equal 2, both \( \sin 2x \) and \( \cos 4x \) must equal 1 simultaneously.

### Step 3: Set up the equations
1. \( \sin 2x = 1 \)
2. \( \cos 4x = 1 \)

### Step 4: Solve \( \sin 2x = 1 \)
The equation \( \sin 2x = 1 \) occurs when:
\[ 2x = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ x = \frac{\pi}{4} + n\pi \]

### Step 5: Solve \( \cos 4x = 1 \)
The equation \( \cos 4x = 1 \) occurs when:
\[ 4x = 2m\pi \quad \text{for } m \in \mathbb{Z} \]
Thus,
\[ x = \frac{m\pi}{2} \]

### Step 6: Find common solutions
Now we need to find values of \( x \) that satisfy both equations:
1. From \( x = \frac{\pi}{4} + n\pi \)
2. From \( x = \frac{m\pi}{2} \)

Set the two equations equal to each other:
\[ \frac{\pi}{4} + n\pi = \frac{m\pi}{2} \]
Multiply through by 4 to eliminate the fractions:
\[ \pi + 4n\pi = 2m\pi \]
Rearranging gives:
\[ 4n + 2m = 1 \]
This equation does not have integer solutions for \( n \) and \( m \) since the left side is even and the right side is odd.

### Step 7: Conclusion
Since there are no integer solutions for \( n \) and \( m \) that satisfy both conditions, we conclude that the original equation \( \sin 2x + \cos 4x = 2 \) has no solutions.

### Final Answer
The solution of the equation \( \sin 2x + \cos 4x = 2 \) is:
\[ \text{No solution} \]

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