The equation `sin^4 x - 2cos^2 x + a^2 = 0` is solveble if

Options

A. `- sqrt3 le a le sqrt3`
B. `-sqrt2 le a le sqrt2`
C. `-1 le a le 1`
D. none of these

Answer

To solve the equation \( \sin^4 x - 2\cos^2 x + a^2 = 0 \) for the condition under which it is solvable, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin^4 x - 2\cos^2 x + a^2 = 0 \]
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite \( \cos^2 x \):
\[ \sin^4 x - 2(1 - \sin^2 x) + a^2 = 0 \]

### Step 2: Simplify the equation
Expanding the equation gives:
\[ \sin^4 x - 2 + 2\sin^2 x + a^2 = 0 \]
Rearranging this, we have:
\[ \sin^4 x + 2\sin^2 x + (a^2 - 2) = 0 \]

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

### Step 4: Identify the discriminant
This is a quadratic equation in \( y \). For the equation to have real solutions, the discriminant must be non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = 2 \), and \( c = a^2 - 2 \). Thus:
\[ D = 2^2 - 4 \cdot 1 \cdot (a^2 - 2) \]
\[ D = 4 - 4(a^2 - 2) = 4 - 4a^2 + 8 = 12 - 4a^2 \]

### Step 5: Set the discriminant greater than or equal to zero
For the quadratic to have real solutions, we require:
\[ 12 - 4a^2 \geq 0 \]
Solving this inequality:
\[ 12 \geq 4a^2 \]
\[ 3 \geq a^2 \]
\[ a^2 \leq 3 \]

### Step 6: Find the range of \( a \)
Taking the square root gives:
\[ -\sqrt{3} \leq a \leq \sqrt{3} \]

### Final Condition
Thus, the equation \( \sin^4 x - 2\cos^2 x + a^2 = 0 \) is solvable if:
\[ a \in [-\sqrt{3}, \sqrt{3}] \]