`cos2 theta+2 costheta` is always

Options

A. greater than `-3/2`
B. less than or equal to `3/2`
C. greater than or equal to `-3/2`
D. none of these

Answer

To find the range of the expression \( \cos 2\theta + 2\cos \theta \), we will follow these steps:

### Step 1: Use the double angle identity for cosine
The double angle identity states that:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
Substituting this into our expression gives:
\[ \cos 2\theta + 2\cos \theta = (2\cos^2 \theta - 1) + 2\cos \theta \]

### Step 2: Simplify the expression
Now, we can combine like terms:
\[ 2\cos^2 \theta + 2\cos \theta - 1 \]
This can be rewritten as:
\[ 2\cos^2 \theta + 2\cos \theta - 1 = 2\cos^2 \theta + 2\cos \theta - 1 \]

### Step 3: Let \( x = \cos \theta \)
To make it easier to analyze, let \( x = \cos \theta \). The expression now becomes:
\[ 2x^2 + 2x - 1 \]

### Step 4: Find the vertex of the quadratic
The quadratic \( 2x^2 + 2x - 1 \) can be analyzed using the vertex formula. The vertex \( x \) of a quadratic \( ax^2 + bx + c \) is given by:
\[ x = -\frac{b}{2a} \]
Here, \( a = 2 \) and \( b = 2 \):
\[ x = -\frac{2}{2 \cdot 2} = -\frac{1}{2} \]

### Step 5: Calculate the maximum or minimum value
Now, substitute \( x = -\frac{1}{2} \) back into the expression to find the minimum value:
\[ 2\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) - 1 \]
Calculating this gives:
\[ 2 \cdot \frac{1}{4} - 1 - 1 = \frac{1}{2} - 1 - 1 = -\frac{3}{2} \]

### Step 6: Determine the range of the expression
Since \( \cos \theta \) ranges from -1 to 1, we need to check the values of the quadratic at these endpoints:
1. For \( x = -1 \):
\[ 2(-1)^2 + 2(-1) - 1 = 2 - 2 - 1 = -1 \]
2. For \( x = 1 \):
\[ 2(1)^2 + 2(1) - 1 = 2 + 2 - 1 = 3 \]

### Conclusion
The minimum value of \( 2\cos^2 \theta + 2\cos \theta - 1 \) is \( -\frac{3}{2} \) and the maximum value is \( 3 \). Therefore, the expression \( \cos 2\theta + 2\cos \theta \) is always:
\[ \text{Greater than or equal to } -\frac{3}{2} \text{ and less than or equal to } 3. \]