Let the inequality `sin ^(2) x+a cos x +a ^(2) ge1+ cos x ` is satisfied `AA x in R,` for `a in (-oo,k_(1)] uu[ k_(2), oo),` then `|k_(1)|+ |k_(2)|=`

Answer

To solve the inequality \( \sin^2 x + a \cos x + a^2 \geq 1 + \cos x \) for all \( x \in \mathbb{R} \) and find the values of \( k_1 \) and \( k_2 \) such that \( a \in (-\infty, k_1] \cup [k_2, \infty) \), we can follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \sin^2 x + a \cos x + a^2 \geq 1 + \cos x \]
We can rearrange this to:
\[ \sin^2 x + a \cos x + a^2 - 1 - \cos x \geq 0 \]

### Step 2: Use the Identity for Sine
Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can substitute:
\[ 1 - \cos^2 x + a \cos x + a^2 - 1 - \cos x \geq 0 \]
This simplifies to:
\[ -\cos^2 x + (a - 1) \cos x + a^2 \geq 0 \]

### Step 3: Rearranging the Terms
Rearranging gives us:
\[ \cos^2 x - (a - 1) \cos x - a^2 \leq 0 \]
This is a quadratic inequality in terms of \( \cos x \).

### Step 4: Determine the Discriminant
For the quadratic \( \cos^2 x - (a - 1) \cos x - a^2 \) to be non-positive for all \( x \), its discriminant must be less than or equal to zero:
\[ D = (a - 1)^2 + 4a^2 \]
Setting the discriminant \( D \leq 0 \):
\[ (a - 1)^2 + 4a^2 \leq 0 \]
This simplifies to:
\[ 5a^2 - 2a + 1 \leq 0 \]

### Step 5: Solve the Quadratic Inequality
To find the roots of the equation \( 5a^2 - 2a + 1 = 0 \), we calculate the discriminant:
\[ D' = (-2)^2 - 4 \cdot 5 \cdot 1 = 4 - 20 = -16 \]
Since the discriminant is negative, the quadratic \( 5a^2 - 2a + 1 \) does not cross the x-axis and is always positive. Therefore, it does not satisfy the inequality.

### Step 6: Analyze the Roots
The inequality \( 5a^2 - 2a + 1 \leq 0 \) has no real solutions, indicating that the inequality \( \sin^2 x + a \cos x + a^2 \geq 1 + \cos x \) is satisfied for \( a \) in the intervals:
\[ (-\infty, k_1] \cup [k_2, \infty) \]
where \( k_1 = -2 \) and \( k_2 = 1 \).

### Step 7: Calculate \( |k_1| + |k_2| \)
Now, we find:
\[ |k_1| + |k_2| = |-2| + |1| = 2 + 1 = 3 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{3} \]