Consider the equaiton k sinx + cos 2x=2k-7 If the equaiton possesses solution then that what is the minimum value of k ?

Options

A. 1
B. 2
C. 4
D. 6

Explanation

To solve the equation \( k \sin x + \cos 2x = 2k - 7 \) and find the minimum value of \( k \) for which the equation possesses a solution, we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \cos 2x = 1 - 2\sin^2 x \). Substituting this into the equation gives:
\[ k \sin x + (1 - 2\sin^2 x) = 2k - 7 \]
This simplifies to:
\[ k \sin x - 2\sin^2 x + 1 = 2k - 7 \]

### Step 2: Rearranging the equation
Rearranging the equation, we get:
\[ -2\sin^2 x + k \sin x + 1 - 2k + 7 = 0 \]
This simplifies to:
\[ -2\sin^2 x + k \sin x + 8 - 2k = 0 \]
Multiplying through by -1 gives:
\[ 2\sin^2 x - k \sin x + (2k - 8) = 0 \]

### Step 3: Identify the quadratic form
This is a quadratic equation in the form \( A \sin^2 x + B \sin x + C = 0 \) where:
- \( A = 2 \)
- \( B = -k \)
- \( C = 2k - 8 \)

### Step 4: Use the discriminant condition
For the quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = B^2 - 4AC \geq 0 \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ (-k)^2 - 4(2)(2k - 8) \geq 0 \]
This simplifies to:
\[ k^2 - 8(2k - 8) \geq 0 \]
\[ k^2 - 16k + 64 \geq 0 \]

### Step 5: Factor the quadratic
Factoring the quadratic:
\[ (k - 8)^2 \geq 0 \]
This inequality holds for all \( k \) and is equal to zero when \( k = 8 \).

### Step 6: Determine the minimum value of \( k \)
The minimum value of \( k \) for which the equation has at least one solution occurs at the point where the discriminant is zero:
\[ k = 8 \]

### Conclusion
Thus, the minimum value of \( k \) for which the equation \( k \sin x + \cos 2x = 2k - 7 \) possesses a solution is:
\[ \boxed{8} \]