Find the smallest positive number p for which the equation `cos (p sinx) = sin ( p cos x)` has a solution when ` x in [0,2pi].`

Answer

To find the smallest positive number \( p \) for which the equation \( \cos(p \sin x) = \sin(p \cos x) \) has a solution when \( x \in [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos(p \sin x) = \sin(p \cos x) \]
Using the identity \( \sin \theta = \cos\left(\frac{\pi}{2} - \theta\right) \), we can rewrite the right-hand side:
\[ \cos(p \sin x) = \cos\left(\frac{\pi}{2} - p \cos x\right) \]

### Step 2: Set the arguments equal
Since \( \cos A = \cos B \) implies \( A = 2n\pi \pm B \) for some integer \( n \), we can set:
\[ p \sin x = 2n\pi \pm \left(\frac{\pi}{2} - p \cos x\right) \]

### Step 3: Rearranging the equation
Rearranging gives us two cases to consider:
1. \( p \sin x = 2n\pi + \frac{\pi}{2} - p \cos x \)
2. \( p \sin x = 2n\pi - \frac{\pi}{2} + p \cos x \)

For simplicity, let's focus on the first case:
\[ p \sin x + p \cos x = 2n\pi + \frac{\pi}{2} \]

### Step 4: Factor out \( p \)
Factoring out \( p \) gives:
\[ p(\sin x + \cos x) = 2n\pi + \frac{\pi}{2} \]

### Step 5: Solve for \( p \)
Thus, we can express \( p \) as:
\[ p = \frac{2n\pi + \frac{\pi}{2}}{\sin x + \cos x} \]

### Step 6: Find the maximum of \( \sin x + \cos x \)
The maximum value of \( \sin x + \cos x \) occurs when \( x = \frac{\pi}{4} \) or \( x = \frac{5\pi}{4} \), and is given by:
\[ \sin x + \cos x = \sqrt{2} \]

### Step 7: Substitute the maximum value
Substituting this into our equation for \( p \):
\[ p = \frac{2n\pi + \frac{\pi}{2}}{\sqrt{2}} \]

### Step 8: Find the smallest positive \( p \)
To find the smallest positive \( p \), we set \( n = 0 \):
\[ p = \frac{\frac{\pi}{2}}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} = \frac{\pi \sqrt{2}}{4} \]

### Final Answer
Thus, the smallest positive number \( p \) for which the equation has a solution is:
\[ \boxed{\frac{\pi \sqrt{2}}{4}} \]