If `cos^2x+cosx=a+2`, then find the value of a for which equation has solution.

Explanation

To solve the equation \( \cos^2 x + \cos x = a + 2 \) and find the value of \( a \) for which the equation has solutions, we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^2 x + \cos x - (a + 2) = 0 \]
This is a quadratic equation in terms of \( \cos x \).

### Step 2: Identify the quadratic form
Let \( y = \cos x \). The equation can be rewritten as:
\[ y^2 + y - (a + 2) = 0 \]
This is a standard quadratic equation of the form \( Ay^2 + By + C = 0 \) where \( A = 1 \), \( B = 1 \), and \( C = -(a + 2) \).

### Step 3: Apply the discriminant condition
For the quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = B^2 - 4AC \geq 0 \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ D = 1^2 - 4 \cdot 1 \cdot (-(a + 2)) \geq 0 \]
This simplifies to:
\[ 1 + 4(a + 2) \geq 0 \]
\[ 1 + 4a + 8 \geq 0 \]
\[ 4a + 9 \geq 0 \]

### Step 4: Solve for \( a \)
Now, we solve the inequality:
\[ 4a \geq -9 \]
\[ a \geq -\frac{9}{4} \]

### Step 5: Consider the range of \( \cos x \)
Since \( \cos x \) has a range of \([-1, 1]\), we need to ensure that the values of \( y = \cos x \) also satisfy this range. The maximum value of \( y \) is 1, which gives us:
\[ y^2 + y - (a + 2) = 0 \quad \text{at } y = 1 \]
Substituting \( y = 1 \):
\[ 1^2 + 1 - (a + 2) = 0 \]
\[ 2 - (a + 2) = 0 \]
\[ -a = 0 \implies a = 0 \]

### Step 6: Combine the conditions
From the discriminant condition, we found \( a \geq -\frac{9}{4} \). From the maximum value condition, we found \( a = 0 \). Thus, the value of \( a \) for which the equation has solutions is:
\[ -\frac{9}{4} \leq a \leq 0 \]

### Final Answer
The value of \( a \) for which the equation has solutions is:
\[ a \in \left[-\frac{9}{4}, 0\right] \]