The equation `2sin^(2)x=8-k(2-sinx)` possesses a solution if

Options

A. `k in [-6,-2]`
B. `k in [-6,2]`
C. `k in [-2,6]`
D. `k in [2,6]`

Answer

To solve the equation \(2\sin^2 x = 8 - k(2 - \sin x)\) and determine the values of \(k\) for which this equation possesses a solution, we can follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 2\sin^2 x = 8 - k(2 - \sin x) \]
Expanding the right-hand side:
\[ 2\sin^2 x = 8 - 2k + k\sin x \]
Rearranging gives:
\[ 2\sin^2 x - k\sin x + (2k - 8) = 0 \]

### Step 2: Identifying the Quadratic Form
This is a quadratic equation in terms of \(\sin x\):
\[ 2\sin^2 x - k\sin x + (2k - 8) = 0 \]
Here, we identify:
- \(a = 2\)
- \(b = -k\)
- \(c = 2k - 8\)

### Step 3: Finding the Discriminant
For the quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (-k)^2 - 4(2)(2k - 8) \]
Calculating this gives:
\[ D = k^2 - 8(2k - 8) = k^2 - 16k + 64 \]
This can be rewritten as:
\[ D = (k - 8)^2 \]

### Step 4: Setting the Discriminant Condition
For the equation to have real solutions, we require:
\[ (k - 8)^2 \geq 0 \]
This condition is always satisfied for all \(k\), since a square is always non-negative.

### Step 5: Considering the Range of \(\sin x\)
Next, we need to ensure that the solutions for \(\sin x\) fall within the valid range of the sine function, which is \([-1, 1]\).

### Step 6: Finding the Values of \(k\)
From the quadratic formula, we have:
\[ \sin x = \frac{k \pm (k - 8)}{4} \]
This simplifies to two cases:
1. \(\sin x = \frac{2k - 8}{4} = \frac{k - 4}{2}\)
2. \(\sin x = \frac{8}{4} = 2\)

The second case is invalid since \(\sin x\) cannot exceed 1. Thus, we focus on the first case:
\[ -1 \leq \frac{k - 4}{2} \leq 1 \]

### Step 7: Solving the Inequalities
1. From \(\frac{k - 4}{2} \geq -1\):
\[ k - 4 \geq -2 \implies k \geq 2 \]
2. From \(\frac{k - 4}{2} \leq 1\):
\[ k - 4 \leq 2 \implies k \leq 6 \]

### Step 8: Final Range
Combining both inequalities, we find:
\[ 2 \leq k \leq 6 \]
Thus, the range of values for \(k\) for which the equation possesses a solution is:
\[ k \in [2, 6] \]