The equation `"sin"^(4) x - (k +2)"sin"^(2) x - (k + 3) = 0` possesses a solution, if

Options

A. `k gt -3`
B. `k lt -2`
C. `-3 le k le -2`
D. `k in Z`

Explanation

To solve the equation \( \sin^4 x - (k + 2) \sin^2 x - (k + 3) = 0 \) for the values of \( k \) for which it possesses a solution, we can follow these steps:

### Step 1: Substitution
Let \( y = \sin^2 x \). Then, the equation can be rewritten as:
\[ y^2 - (k + 2)y - (k + 3) = 0 \]

### Step 2: Identify Coefficients
In the quadratic equation \( ay^2 + by + c = 0 \), we identify:
- \( a = 1 \)
- \( b = -(k + 2) \)
- \( c = -(k + 3) \)

### Step 3: Use the Quadratic Formula
The roots of the quadratic equation can be found using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ y = \frac{k + 2 \pm \sqrt{(k + 2)^2 - 4 \cdot 1 \cdot (-(k + 3))}}{2 \cdot 1} \]

### Step 4: Simplify the Discriminant
Now, simplify the expression under the square root (the discriminant):
\[ (k + 2)^2 + 4(k + 3) = k^2 + 4k + 4 + 4k + 12 = k^2 + 8k + 16 = (k + 4)^2 \]
Thus, the roots become:
\[ y = \frac{k + 2 \pm (k + 4)}{2} \]

### Step 5: Calculate the Roots
Calculating the two possible roots:
1. For the positive sign:
\[ y_1 = \frac{k + 2 + (k + 4)}{2} = \frac{2k + 6}{2} = k + 3 \]
2. For the negative sign:
\[ y_2 = \frac{k + 2 - (k + 4)}{2} = \frac{-2}{2} = -1 \]

### Step 6: Determine Validity of Roots
Since \( y = \sin^2 x \), it must satisfy:
\[ 0 \leq y \leq 1 \]
From \( y_1 = k + 3 \):
- For \( k + 3 \geq 0 \) implies \( k \geq -3 \)
- For \( k + 3 \leq 1 \) implies \( k \leq -2 \)

From \( y_2 = -1 \):
- This value is not valid since \( \sin^2 x \) cannot be negative.

### Step 7: Final Inequality
Thus, combining the inequalities, we find:
\[ -3 \leq k \leq -2 \]

### Conclusion
The equation \( \sin^4 x - (k + 2) \sin^2 x - (k + 3) = 0 \) possesses a solution if:
\[ k \in [-3, -2] \]