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

Options

A. `[1-sqrt(5)/(2),(1+sqrt(5))/(2)]`
B. `[infty,(1+sqrt(5))/(2)]`
C. `[(1-sqrt(5))/(2),infty]`
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 using the Pythagorean identity
We know that \( \cos^2 x = 1 - \sin^2 x \). Substitute this into the equation:

\[ \sin^4 x - 2(1 - \sin^2 x) + a^2 = 0 \]

### Step 2: Simplify the equation
Distributing the \(-2\) 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 \( y = \sin^2 x \)
Now, let \( y = \sin^2 x \). The equation becomes a quadratic in \( y \):

\[ y^2 + 2y + (a^2 - 2) = 0 \]

### Step 4: Use the quadratic formula
The roots of the quadratic equation \( ay^2 + by + c = 0 \) are given by:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, \( a = 1 \), \( b = 2 \), and \( c = a^2 - 2 \). Thus, the roots are:

\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (a^2 - 2)}}{2 \cdot 1} \]

This simplifies to:

\[ y = \frac{-2 \pm \sqrt{4 - 4(a^2 - 2)}}{2} \]

### Step 5: Simplify the expression under the square root
This can be further simplified:

\[ y = \frac{-2 \pm \sqrt{4 - 4a^2 + 8}}{2} = \frac{-2 \pm \sqrt{12 - 4a^2}}{2} \]

\[ y = -1 \pm \sqrt{3 - a^2} \]

### Step 6: Determine the conditions for \( y \)
Since \( y = \sin^2 x \) must lie in the interval \([0, 1]\), we need to analyze the two cases:

1. **Case 1**:
\[ -1 + \sqrt{3 - a^2} \geq 0 \]
This implies:
\[ \sqrt{3 - a^2} \geq 1 \implies 3 - a^2 \geq 1 \implies a^2 \leq 2 \]

2. **Case 2**:
\[ -1 - \sqrt{3 - a^2} \leq 1 \]
This is always satisfied since \(-1 - \sqrt{3 - a^2}\) will always be less than or equal to 1.

### Step 7: Combine the conditions
From Case 1, we have:

\[ a^2 \leq 2 \]

Thus, the equation \( \sin^4 x - 2 \cos^2 x + a^2 = 0 \) is solvable if:

\[ a^2 \leq 2 \]

### Final Answer
The equation is solvable if \( a^2 \leq 2 \).

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