If `sin^(2)x` - (m - 3)sin x + m=0 "has real roots then which of the following is/are correct"

Options

A. Number of roots of equation in `[-2pi,2pi]is 2 for m=1`
B. Number of roots of equation in `[-2pi,2pi]is 1 for m=9`
C. Roots are real for `m le 1`
D. If sin x =t then `t^(2)-(m-3)t+m=0` can not have both roots in [-1,1]`

Answer

To solve the equation \( \sin^2 x - (m - 3) \sin x + m = 0 \) for real roots, we can follow these steps:

### Step 1: Substitute \( \sin x \) with \( t \)
Let \( t = \sin x \). The equation becomes:
\[ t^2 - (m - 3)t + m = 0 \]

### Step 2: Identify coefficients
In this quadratic equation, we have:
- \( a = 1 \)
- \( b = -(m - 3) \)
- \( c = m \)

### Step 3: Apply the discriminant condition
For the quadratic equation to have real roots, the discriminant \( D \) must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-(m - 3))^2 - 4 \cdot 1 \cdot m \geq 0 \]
This simplifies to:
\[ (m - 3)^2 - 4m \geq 0 \]

### Step 4: Expand and simplify the inequality
Expanding \( (m - 3)^2 \):
\[ m^2 - 6m + 9 - 4m \geq 0 \]
Combining like terms:
\[ m^2 - 10m + 9 \geq 0 \]

### Step 5: Factor the quadratic inequality
Now we need to factor \( m^2 - 10m + 9 \):
\[ (m - 1)(m - 9) \geq 0 \]

### Step 6: Determine the intervals
To solve the inequality \( (m - 1)(m - 9) \geq 0 \), we find the critical points where the expression equals zero:
- \( m = 1 \)
- \( m = 9 \)

The sign of the product changes at these points. We test intervals:
1. \( m < 1 \): Both factors are negative, so the product is positive.
2. \( 1 < m < 9 \): One factor is negative and the other is positive, so the product is negative.
3. \( m > 9 \): Both factors are positive, so the product is positive.

Thus, the solution to the inequality is:
\[ m \leq 1 \quad \text{or} \quad m \geq 9 \]

### Step 7: Conclusion
The values of \( m \) for which the original equation has real roots are:
\[ m \in (-\infty, 1] \cup [9, \infty) \]