The equation `"sin"^(4) x -2 "cos"^(2) x + a^(2) =0` is solvable if

Options

A. `-sqrt(3) le a le sqrt(3)`
B. `-sqrt(2) le a le sqrt(2)`
C. `-1 le a le 1`
D. none of these

Answer

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

### Step 1: Rewrite the Equation
The given equation is:
\[ \sin^4 x - 2 \cos^2 x + a^2 = 0 \]
We can express \( \cos^2 x \) in terms of \( \sin^2 x \):
\[ \cos^2 x = 1 - \sin^2 x \]
Substituting this into the equation gives:
\[ \sin^4 x - 2(1 - \sin^2 x) + a^2 = 0 \]

### Step 2: Simplify the Equation
Expanding the equation:
\[ \sin^4 x - 2 + 2 \sin^2 x + a^2 = 0 \]
Rearranging it, 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: Determine the Discriminant
For this quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (a^2 - 2) \geq 0 \]
Calculating the discriminant:
\[ D = 4 - 4(a^2 - 2) \geq 0 \]
This simplifies to:
\[ 4 - 4a^2 + 8 \geq 0 \]
\[ 12 - 4a^2 \geq 0 \]
Dividing by 4:
\[ 3 - a^2 \geq 0 \]
Rearranging gives:
\[ a^2 \leq 3 \]

### Step 5: Find the Interval for \( a \)
Taking the square root of both sides, we find:
\[ -\sqrt{3} \leq a \leq \sqrt{3} \]

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