If the equation `cos 2x+alpha sinx=2alpha-7` has a solution. Then range of `alpha` is (A) `R` (B) `[1,4]` (C) `[3,7]` (D) `[2,6]`

Options

A. `R`
B. `[1,4]`
C. `[3,7]`
D. `[2,6]`

Answer

To find the range of \(\alpha\) for which the equation \(\cos 2x + \alpha \sin x = 2\alpha - 7\) has a solution, we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \(\cos 2x = 1 - 2\sin^2 x\). Thus, we can rewrite the equation as:
\[ 1 - 2\sin^2 x + \alpha \sin x = 2\alpha - 7 \]

### Step 2: Rearranging the equation
Rearranging gives us:
\[ -2\sin^2 x + \alpha \sin x + 1 - 2\alpha + 7 = 0 \]
This simplifies to:
\[ -2\sin^2 x + \alpha \sin x + 8 - 2\alpha = 0 \]

### Step 3: Multiply through by -1
To make the quadratic easier to work with, we can multiply the entire equation by -1:
\[ 2\sin^2 x - \alpha \sin x + (2\alpha - 8) = 0 \]

### Step 4: Identify coefficients for the quadratic equation
This is a quadratic equation in terms of \(\sin x\):
- \(a = 2\)
- \(b = -\alpha\)
- \(c = 2\alpha - 8\)

### Step 5: Calculate the discriminant
For the quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values:
\[ (-\alpha)^2 - 4(2)(2\alpha - 8) \geq 0 \]
This simplifies to:
\[ \alpha^2 - 8(2\alpha - 8) \geq 0 \]
\[ \alpha^2 - 16\alpha + 64 \geq 0 \]

### Step 6: Factor the quadratic
We can factor the quadratic:
\[ (\alpha - 8)^2 \geq 0 \]
This inequality holds for all \(\alpha\), but we need to find the values of \(\alpha\) that keep \(\sin x\) within its valid range, which is \([-1, 1]\).

### Step 7: Set the bounds for \(\sin x\)
From the quadratic equation \(2\sin^2 x - \alpha \sin x + (2\alpha - 8) = 0\), we need to ensure that the roots (values of \(\sin x\)) are within the interval \([-1, 1]\).

### Step 8: Solve for the bounds of \(\alpha\)
Using the quadratic formula:
\[ \sin x = \frac{\alpha \pm \sqrt{(\alpha - 8)^2}}{4} \]
The roots are:
\[ \sin x = \frac{\alpha + (\alpha - 8)}{4} = \frac{2\alpha - 8}{4} = \frac{\alpha - 4}{2} \]
\[ \sin x = \frac{\alpha - (\alpha - 8)}{4} = \frac{8}{4} = 2 \]
Since \(\sin x\) cannot be greater than 1, we discard the second root.

Now, we need:
\[ -1 \leq \frac{\alpha - 4}{2} \leq 1 \]

### Step 9: Solve the inequalities
1. From \(\frac{\alpha - 4}{2} \geq -1\):
\[ \alpha - 4 \geq -2 \implies \alpha \geq 2 \]

2. From \(\frac{\alpha - 4}{2} \leq 1\):
\[ \alpha - 4 \leq 2 \implies \alpha \leq 6 \]

### Conclusion
Thus, the range of \(\alpha\) for which the equation has a solution is:
\[ \alpha \in [2, 6] \]

### Final Answer
The correct option is (D) \([2, 6]\).