Write the general solution of the following differential equations
`(dy)/(dx)= (1-cos 2x)/(1+cos 2y)`

Answer

To solve the differential equation \(\frac{dy}{dx} = \frac{1 - \cos 2x}{1 + \cos 2y}\), we will separate the variables and integrate both sides.

### Step 1: Separate the variables
We start by rearranging the equation to separate the variables \(y\) and \(x\):
\[ (1 + \cos 2y) dy = (1 - \cos 2x) dx \]

### Step 2: Integrate both sides
Now we will integrate both sides:
\[ \int (1 + \cos 2y) dy = \int (1 - \cos 2x) dx \]

### Step 3: Compute the integrals
Now we compute the integrals on both sides.

**Left side:**
\[ \int (1 + \cos 2y) dy = \int 1 \, dy + \int \cos 2y \, dy = y + \frac{\sin 2y}{2} + C_1 \]

**Right side:**
\[ \int (1 - \cos 2x) dx = \int 1 \, dx - \int \cos 2x \, dx = x - \frac{\sin 2x}{2} + C_2 \]

### Step 4: Combine the results
Setting the results from both integrals equal to each other gives us:
\[ y + \frac{\sin 2y}{2} = x - \frac{\sin 2x}{2} + C \]
where \(C = C_2 - C_1\) is a constant.

### Step 5: Rearranging the equation
Rearranging the equation, we can express it as:
\[ 2y - x + \sin 2y + \sin 2x = 2C \]

### Step 6: Final form
Let \(k = 2C\), we can write the general solution as:
\[ 2y - x + \sin 2y + \sin 2x = k \]

### Summary of the solution
Thus, the general solution of the differential equation is:
\[ 2y - x + \sin 2y + \sin 2x = k \]