If `sin^2theta=x^2-3x+3` is meaningful, then find the values of x.

Explanation

To solve the problem, we need to determine the values of \( x \) for which the expression \( \sin^2 \theta = x^2 - 3x + 3 \) is meaningful. This means we need to ensure that the value of \( \sin^2 \theta \) lies within the range of valid values for the sine function, which is between 0 and 1 (inclusive).

### Step 1: Set up the inequalities
Since \( \sin^2 \theta \) must be between 0 and 1, we can set up the following inequalities:
\[ 0 \leq x^2 - 3x + 3 \leq 1 \]

### Step 2: Solve the first inequality
First, we solve the inequality:
\[ x^2 - 3x + 3 \geq 0 \]
To determine when this quadratic expression is greater than or equal to zero, we can find its discriminant:
\[ D = b^2 - 4ac = (-3)^2 - 4(1)(3) = 9 - 12 = -3 \]
Since the discriminant is negative, the quadratic does not have real roots and is always positive (as the coefficient of \( x^2 \) is positive). Thus, this inequality holds for all real numbers \( x \).

### Step 3: Solve the second inequality
Next, we solve the second inequality:
\[ x^2 - 3x + 3 \leq 1 \]
Rearranging gives:
\[ x^2 - 3x + 2 \leq 0 \]
Now, we can factor the quadratic:
\[ (x - 1)(x - 2) \leq 0 \]

### Step 4: Determine the intervals
To find the intervals where this inequality holds, we can analyze the sign of the product \( (x - 1)(x - 2) \):
- The roots are \( x = 1 \) and \( x = 2 \).
- Testing intervals:
- For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 1)(0 - 2) = 2 > 0 \) (not valid)
- For \( 1 < x < 2 \): Choose \( x = 1.5 \) → \( (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0 \) (valid)
- For \( x > 2 \): Choose \( x = 3 \) → \( (3 - 1)(3 - 2) = 2 > 0 \) (not valid)

Thus, the solution to the inequality \( (x - 1)(x - 2) \leq 0 \) is:
\[ 1 \leq x \leq 2 \]

### Step 5: Conclusion
The values of \( x \) for which \( \sin^2 \theta = x^2 - 3x + 3 \) is meaningful are:
\[ x \in [1, 2] \]