To find the number of roots of the equation \( x^2 + \sin^2 x = 1 \) in the closed interval \([0, \frac{\pi}{2}]\), we can follow these steps:
### Step 1: Rearranging the Equation
We start with the equation:
\[
x^2 + \sin^2 x = 1
\]
Rearranging gives us:
\[
x^2 = 1 - \sin^2 x
\]
Using the Pythagorean identity, we know that:
\[
1 - \sin^2 x = \cos^2 x
\]
Thus, we can rewrite the equation as:
\[
x^2 = \cos^2 x
\]
### Step 2: Setting Up the New Equation
We can express this as:
\[
x^2 - \cos^2 x = 0
\]
This can be factored or rewritten as:
\[
x^2 = \cos^2 x
\]
### Step 3: Analyzing the Functions
Next, we will analyze the functions \( y_1 = x^2 \) and \( y_2 = \cos^2 x \) over the interval \([0, \frac{\pi}{2}]\).
- The function \( y_1 = x^2 \) is a parabola that starts at \( (0, 0) \) and increases to \( \left(\frac{\pi}{2}\right)^2 \) at \( x = \frac{\pi}{2} \).
- The function \( y_2 = \cos^2 x \) starts at \( \cos^2(0) = 1 \) and decreases to \( \cos^2\left(\frac{\pi}{2}\right) = 0 \).
### Step 4: Finding Intersection Points
To find the number of roots, we need to determine how many times these two curves intersect in the interval \([0, \frac{\pi}{2}]\).
1. At \( x = 0 \):
\[
y_1 = 0^2 = 0, \quad y_2 = \cos^2(0) = 1
\]
(The parabola is below the cosine curve.)
2. At \( x = \frac{\pi}{2} \):
\[
y_1 = \left(\frac{\pi}{2}\right)^2, \quad y_2 = \cos^2\left(\frac{\pi}{2}\right) = 0
\]
(The parabola is above the cosine curve.)
Since \( y_1 \) starts below \( y_2 \) and ends above \( y_2 \), and both functions are continuous, by the Intermediate Value Theorem, there must be at least one intersection point in the interval \((0, \frac{\pi}{2})\).
### Step 5: Checking for More Intersections
To check if there are more than one intersection, we can analyze the slopes:
- The derivative of \( y_1 = x^2 \) is \( 2x \), which is increasing.
- The derivative of \( y_2 = \cos^2 x \) is \( -2\cos x \sin x \) (using the chain rule), which is decreasing in the interval \([0, \frac{\pi}{2}]\).
Since \( y_1 \) is increasing and \( y_2 \) is decreasing, they can intersect at most once in the interval.
### Conclusion
Thus, we conclude that there is exactly **one root** of the equation \( x^2 + \sin^2 x = 1 \) in the interval \([0, \frac{\pi}{2}]\).
The answer is:
\[
\text{Number of roots} = 1
\]
