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

Options

A. 2
B. 1
C. 3
D. none of these

Explanation

To solve the equation \( \cos(p \sin x) = \sin(p \cos x) \) and find the smallest positive number \( p \) for which there exists a solution \( x \in [0, 2\pi] \), 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 \( \cos A = \sin\left(\frac{\pi}{2} - A\right) \), we can rewrite the left-hand side:
\[ \sin\left(\frac{\pi}{2} - p \sin x\right) = \sin(p \cos x) \]

### Step 2: Set the arguments equal
Since \( \sin A = \sin B \) implies that \( A = B + 2k\pi \) or \( A = \pi - B + 2k\pi \) for some integer \( k \), we can set the arguments equal:
1. \( \frac{\pi}{2} - p \sin x = p \cos x + 2k\pi \)
2. \( \frac{\pi}{2} - p \sin x = \pi - p \cos x + 2k\pi \)

### Step 3: Analyze the first case
From the first case, rearranging gives:
\[ p \sin x + p \cos x = \frac{\pi}{2} - 2k\pi \]
Factoring out \( p \):
\[ p(\sin x + \cos x) = \frac{\pi}{2} - 2k\pi \]
Thus,
\[ p = \frac{\frac{\pi}{2} - 2k\pi}{\sin x + \cos x} \]

### Step 4: Analyze the second case
From the second case, rearranging gives:
\[ p \sin x + p \cos x = \pi - \frac{\pi}{2} + 2k\pi \]
This simplifies to:
\[ p(\sin x + \cos x) = \frac{\pi}{2} + 2k\pi \]
Thus,
\[ p = \frac{\frac{\pi}{2} + 2k\pi}{\sin x + \cos x} \]

### Step 5: Find the maximum value of \( \sin x + \cos x \)
The expression \( \sin x + \cos x \) reaches its maximum value of \( \sqrt{2} \) when \( x = \frac{\pi}{4} + n\pi \) for \( n \in \mathbb{Z} \).

### Step 6: Substitute the maximum value
Substituting \( \sin x + \cos x = \sqrt{2} \) into our expressions for \( p \):
1. From the first case:
\[ p = \frac{\frac{\pi}{2} - 2k\pi}{\sqrt{2}} \]
2. From the second case:
\[ p = \frac{\frac{\pi}{2} + 2k\pi}{\sqrt{2}} \]

### Step 7: Find the smallest positive \( p \)
To find the smallest positive \( p \), we can set \( k = 0 \):
1. From the first case:
\[ p = \frac{\frac{\pi}{2}}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} = \frac{\pi \sqrt{2}}{4} \]
2. From the second case:
\[ p = \frac{\frac{\pi}{2}}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} = \frac{\pi \sqrt{2}}{4} \]

### Conclusion
Thus, the smallest positive number \( p \) for which the equation has a solution in the interval \( [0, 2\pi] \) is:
\[ p = \frac{\pi \sqrt{2}}{4} \]