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 lek le-2`
D. k is any(+ve) value

Answer

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

### Step 1: Substitute \( y = \sin^2 x \)

Let \( y = \sin^2 x \). Then, the equation becomes:
\[ y^2 - (k + 2)y - (k + 3) = 0 \]

### Step 2: Identify the coefficients

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

### Step 3: Apply the quadratic formula

The quadratic formula is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

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

Calculate the discriminant:
\[ (k + 2)^2 + 4(k + 3) = k^2 + 4k + 4 + 4k + 12 = k^2 + 8k + 16 \]
This can be rewritten as:
\[ (k + 4)^2 \]

### Step 5: Substitute back into the formula

Now substituting back into the quadratic formula:
\[ y = \frac{k + 2 \pm (k + 4)}{2} \]

### Step 6: Calculate the two possible values of \( y \)

1. For the positive case:
\[ y_1 = \frac{(k + 2) + (k + 4)}{2} = \frac{2k + 6}{2} = k + 3 \]

2. For the negative case:
\[ y_2 = \frac{(k + 2) - (k + 4)}{2} = \frac{-2}{2} = -1 \]

### Step 7: Analyze the values of \( y \)

Since \( y = \sin^2 x \), it must be in the range \( [0, 1] \). Therefore, we analyze the two values obtained:
- \( y_1 = k + 3 \)
- \( y_2 = -1 \) (not valid since \( \sin^2 x \) cannot be negative)

Thus, we need \( k + 3 \) to be within the range \( [0, 1] \):
\[ 0 \leq k + 3 \leq 1 \]

### Step 8: Solve the inequalities

1. From \( k + 3 \geq 0 \):
\[ k \geq -3 \]

2. From \( k + 3 \leq 1 \):
\[ k \leq -2 \]

### Final Result

Combining both inequalities, we find:
\[ -3 \leq k \leq -2 \]

Thus, the equation possesses a solution if \( k \) is in the interval \( [-3, -2] \).