If `y+siny=cosx`, then find the values of 'y' for which `dy/dx` is valid.

Answer

To solve the problem, we start with the equation given:

\[ y + \sin y = \cos x. \]

We need to find the values of \( y \) for which \( \frac{dy}{dx} \) is valid. To do this, we will differentiate both sides of the equation with respect to \( x \).

### Step 1: Differentiate both sides

Differentiating the left side, we apply the chain rule:

\[ \frac{d}{dx}(y + \sin y) = \frac{dy}{dx} + \cos y \frac{dy}{dx} = (1 + \cos y) \frac{dy}{dx}. \]

For the right side, we differentiate \( \cos x \):

\[ \frac{d}{dx}(\cos x) = -\sin x. \]

So, we have:

\[ (1 + \cos y) \frac{dy}{dx} = -\sin x. \]

### Step 2: Solve for \( \frac{dy}{dx} \)

Now, we can isolate \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{-\sin x}{1 + \cos y}. \]

### Step 3: Determine when \( \frac{dy}{dx} \) is valid

For \( \frac{dy}{dx} \) to be valid, the denominator \( 1 + \cos y \) must not be equal to zero:

\[ 1 + \cos y \neq 0. \]

This implies:

\[ \cos y \neq -1. \]

### Step 4: Find values of \( y \)

The condition \( \cos y = -1 \) occurs at:

\[ y = (2n + 1)\pi, \quad n \in \mathbb{Z}. \]

Thus, \( y \) must not take the values:

\[ y \neq (2n + 1)\pi, \quad n \in \mathbb{Z}. \]

### Step 5: Additional conditions from the equation

Next, we analyze the original equation \( y + \sin y = \cos x \). Since \( \cos x \) ranges from -1 to 1, we must also ensure that:

\[ -1 \leq y + \sin y \leq 1. \]

This means we need to check the range of \( y + \sin y \).

### Conclusion

Combining both conditions, we find:

1. \( y \neq (2n + 1)\pi \) for \( n \in \mathbb{Z} \) (to avoid division by zero).
2. \( y + \sin y \) must lie between -1 and 1.

Thus, the values of \( y \) for which \( \frac{dy}{dx} \) is valid are all real numbers except those of the form \( (2n + 1)\pi \).