The number of solutions that the equation `sin (cos (sin x))= cos (sin (cos x))` has in `[0, (pi)/(2)],` is` "_____"`

Answer

To find the number of solutions for the equation \( \sin(\cos(\sin x)) = \cos(\sin(\cos x)) \) in the interval \([0, \frac{\pi}{2}]\), we can follow these steps:

### Step 1: Define the Function
Let \( f(x) = \sin(\cos(\sin x)) - \cos(\sin(\cos x)) \). We want to find the number of solutions to \( f(x) = 0 \) in the interval \([0, \frac{\pi}{2}]\).

### Step 2: Analyze the Function at the Endpoints
Evaluate \( f(0) \) and \( f\left(\frac{\pi}{2}\right) \):
- For \( x = 0 \):
\[ f(0) = \sin(\cos(\sin(0))) - \cos(\sin(\cos(0))) = \sin(\cos(0)) - \cos(\sin(1)) = \sin(1) - \cos(1) \]
Since both \( \sin(1) \) and \( \cos(1) \) are positive, we need to check their values to determine the sign of \( f(0) \).

- For \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \sin(\cos(\sin\left(\frac{\pi}{2}\right))) - \cos(\sin(\cos\left(\frac{\pi}{2}\right))) = \sin(\cos(1)) - \cos(0) = \sin(\cos(1)) - 1 \]
Since \( \sin(\cos(1)) \) is less than or equal to 1, \( f\left(\frac{\pi}{2}\right) < 0 \).

### Step 3: Determine the Behavior of the Function
Next, we need to analyze the derivative \( f'(x) \) to understand the behavior of \( f(x) \):
- Calculate \( f'(x) \) using the chain rule:
\[ f'(x) = \cos(\cos(\sin x)) \cdot (-\sin(\sin x) \cdot \cos x) + \sin(\sin(\cos x)) \cdot (-\cos(\cos x) \cdot \sin x) \]
This expression is complex, but we can observe that both \( \sin x \) and \( \cos x \) are positive in the interval \([0, \frac{\pi}{2}]\).

### Step 4: Analyze the Sign of the Derivative
Since \( \sin(\sin(\cos x)) \) and \( \cos(\cos(\sin x)) \) are both bounded between -1 and 1, we can conclude that \( f'(x) \) is negative in the interval \([0, \frac{\pi}{2}]\). This means that \( f(x) \) is a decreasing function.

### Step 5: Conclusion on the Number of Solutions
Since \( f(0) > 0 \) and \( f\left(\frac{\pi}{2}\right) < 0 \), and given that \( f(x) \) is continuous and decreasing, by the Intermediate Value Theorem, there must be exactly one root in the interval \([0, \frac{\pi}{2}]\).

Thus, the number of solutions to the equation \( \sin(\cos(\sin x)) = \cos(\sin(\cos x)) \) in the interval \([0, \frac{\pi}{2}]\) is:

\[ \boxed{1} \]