If the equation `2cos^(2)x+cosx-a=0` has solutions, then a can be

Options

A. `(-1)/(4)`
B. `(-1)/(8)`
C. `2`
D. `5`

Answer

To solve the equation \(2\cos^2 x + \cos x - a = 0\) for the values of \(a\) that allow solutions, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2\cos^2 x + \cos x - a = 0 \]
Let \(t = \cos x\). Then, we can rewrite the equation as:
\[ 2t^2 + t - a = 0 \]

### Step 2: Identify coefficients
In the quadratic equation \(2t^2 + t - a = 0\), we identify the coefficients:
- \(A = 2\)
- \(B = 1\)
- \(C = -a\)

### Step 3: Use the quadratic formula
The solutions for \(t\) can be found using the quadratic formula:
\[ t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-a)}}{2 \cdot 2} \]
This simplifies to:
\[ t = \frac{-1 \pm \sqrt{1 + 8a}}{4} \]

### Step 4: Determine conditions for real solutions
For \(t\) (which is \(\cos x\)) to have real solutions, the expression under the square root must be non-negative:
\[ 1 + 8a \geq 0 \]
This implies:
\[ 8a \geq -1 \quad \Rightarrow \quad a \geq -\frac{1}{8} \]

### Step 5: Find the range of \(t\)
Since \(t = \cos x\), we know that \(-1 \leq t \leq 1\). Thus, we need to ensure that:
\[ -1 \leq \frac{-1 \pm \sqrt{1 + 8a}}{4} \leq 1 \]

### Step 6: Solve for the upper bound
First, consider the upper bound:
\[ \frac{-1 + \sqrt{1 + 8a}}{4} \leq 1 \]
Multiplying both sides by 4:
\[ -1 + \sqrt{1 + 8a} \leq 4 \]
This leads to:
\[ \sqrt{1 + 8a} \leq 5 \]
Squaring both sides:
\[ 1 + 8a \leq 25 \quad \Rightarrow \quad 8a \leq 24 \quad \Rightarrow \quad a \leq 3 \]

### Step 7: Combine results
Combining the results from steps 4 and 6, we find:
\[ -\frac{1}{8} \leq a \leq 3 \]

### Conclusion
Thus, the values of \(a\) for which the equation \(2\cos^2 x + \cos x - a = 0\) has solutions are:
\[ a \in \left[-\frac{1}{8}, 3\right] \]