Determine all value of 'a' for which the equation ` cos^(4) x-(a+2) cos^(2)x-(a+3)=0`, possess solution.

Answer

To solve the equation \( \cos^4 x - (a + 2) \cos^2 x - (a + 3) = 0 \) for the values of \( a \) for which it possesses solutions, we can follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ \cos^4 x - (a + 2) \cos^2 x - (a + 3) = 0 \]
We can express \( \cos^4 x \) as \( (\cos^2 x)^2 \). Let \( t = \cos^2 x \). Thus, the equation becomes:
\[ t^2 - (a + 2)t - (a + 3) = 0 \]

### Step 2: Identify the Quadratic Form
This is a quadratic equation in \( t \):
\[ t^2 - (a + 2)t - (a + 3) = 0 \]
where \( a \) is a parameter.

### Step 3: Apply the Quadratic Formula
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -(a + 2) \), and \( c = -(a + 3) \):
\[ t = \frac{(a + 2) \pm \sqrt{(a + 2)^2 - 4 \cdot 1 \cdot (-(a + 3))}}{2 \cdot 1} \]

### Step 4: Simplify the Discriminant
Calculating the discriminant:
\[ D = (a + 2)^2 + 4(a + 3) \]
\[ = a^2 + 4a + 4 + 4a + 12 = a^2 + 8a + 16 = (a + 4)^2 \]
Since the discriminant \( D \) is a perfect square, the quadratic equation will have real solutions for all \( a \).

### Step 5: Find the Roots
Now substituting back into the quadratic formula:
\[ t = \frac{(a + 2) \pm (a + 4)}{2} \]
This gives us two cases:
1. \( t = \frac{(a + 2) + (a + 4)}{2} = \frac{2a + 6}{2} = a + 3 \)
2. \( t = \frac{(a + 2) - (a + 4)}{2} = \frac{-2}{2} = -1 \)

### Step 6: Analyze the Values of \( t \)
Since \( t = \cos^2 x \), it must be in the range \( [0, 1] \):
- For \( t = a + 3 \):
\[ 0 \leq a + 3 \leq 1 \]
This simplifies to:
\[ -3 \leq a \leq -2 \]

- For \( t = -1 \):
This is not possible since \( \cos^2 x \) cannot be negative.

### Conclusion
Thus, the values of \( a \) for which the equation has solutions are:
\[ \boxed{[-3, -2]} \]