The set of all values of `lambda` for which the equation `cos^2 2x-2 sin^4 x- 2cos^2 x`=`lamda` has a real solution x, is

Options

A. `[-1,-1/2]`
B. `[-2,-3/2]`
C. `[-2,-1]`
D. `[-3/2,-1]`

Answer

To solve the equation \( \cos^2 2x - 2 \sin^4 x - 2 \cos^2 x = \lambda \) for the set of all values of \( \lambda \) for which there exists a real solution \( x \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^2 2x - 2 \sin^4 x - 2 \cos^2 x = \lambda \]

### Step 2: Use trigonometric identities
Recall that \( \cos 2x = 1 - 2 \sin^2 x \). Thus, we can express \( \cos^2 2x \) as:
\[ \cos^2 2x = (1 - 2 \sin^2 x)^2 = 1 - 4 \sin^2 x + 4 \sin^4 x \]

### Step 3: Substitute into the equation
Now substitute \( \cos^2 2x \) into the equation:
\[ 1 - 4 \sin^2 x + 4 \sin^4 x - 2 \sin^4 x - 2 \cos^2 x = \lambda \]

### Step 4: Simplify the equation
Using \( \cos^2 x = 1 - \sin^2 x \), we can rewrite the equation:
\[ 1 - 4 \sin^2 x + 2 \sin^4 x - 2(1 - \sin^2 x) = \lambda \]
This simplifies to:
\[ 1 - 4 \sin^2 x + 2 \sin^4 x - 2 + 2 \sin^2 x = \lambda \]
Combining like terms gives:
\[ 2 \sin^4 x - 2 \sin^2 x - 1 = \lambda \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ 2 \sin^4 x - 2 \sin^2 x - (\lambda + 1) = 0 \]

### Step 6: Let \( y = \sin^2 x \)
Let \( y = \sin^2 x \). Then the equation becomes:
\[ 2y^2 - 2y - (\lambda + 1) = 0 \]

### Step 7: Find the discriminant
For this quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 2 \cdot (-(\lambda + 1)) \geq 0 \]
Calculating the discriminant:
\[ 4 + 8(\lambda + 1) \geq 0 \]
This simplifies to:
\[ 8\lambda + 12 \geq 0 \]
Thus:
\[ \lambda \geq -\frac{3}{2} \]

### Step 8: Determine the range of \( y \)
Since \( y = \sin^2 x \) must be in the range [0, 1], we need to check the values of \( \lambda \) that keep \( y \) within this range.

### Step 9: Find the maximum value of \( y \)
The maximum value of \( y \) occurs when the quadratic has real solutions. The maximum value of \( y \) can be found by evaluating the vertex of the quadratic:
\[ y = \frac{-b}{2a} = \frac{2}{4} = \frac{1}{2} \]
Substituting \( y = \frac{1}{2} \) into the quadratic:
\[ 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) - (\lambda + 1) = 0 \]
This gives:
\[ \frac{1}{2} - 1 - (\lambda + 1) = 0 \]
Thus:
\[ -\lambda - \frac{3}{2} = 0 \Rightarrow \lambda = -\frac{3}{2} \]

### Step 10: Conclusion
The values of \( \lambda \) for which the equation has real solutions are:
\[ \lambda \in \left[-\frac{3}{2}, 1\right] \]