`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 solve the problem of finding the range of the expression \( \cos(2\theta) + 2\cos(\theta) \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ \cos(2\theta) + 2\cos(\theta) \]
Using the double angle identity for cosine, we know that:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
Substituting this into our expression gives:
\[ (2\cos^2(\theta) - 1) + 2\cos(\theta) \]

### Step 2: Simplify the Expression
Now, we can simplify the expression:
\[ 2\cos^2(\theta) + 2\cos(\theta) - 1 \]

### Step 3: Complete the Square
To make it easier to analyze, we can complete the square. We rewrite the quadratic in terms of \( \cos(\theta) \):
\[ 2\left(\cos^2(\theta) + \cos(\theta)\right) - 1 \]
Now, to complete the square for \( \cos^2(\theta) + \cos(\theta) \):
\[ \cos^2(\theta) + \cos(\theta) = \left(\cos(\theta) + \frac{1}{2}\right)^2 - \frac{1}{4} \]
Substituting this back into our expression gives:
\[ 2\left(\left(\cos(\theta) + \frac{1}{2}\right)^2 - \frac{1}{4}\right) - 1 \]
This simplifies to:
\[ 2\left(\cos(\theta) + \frac{1}{2}\right)^2 - \frac{1}{2} - 1 = 2\left(\cos(\theta) + \frac{1}{2}\right)^2 - \frac{3}{2} \]

### Step 4: Analyze the Expression
The term \( \left(\cos(\theta) + \frac{1}{2}\right)^2 \) is always non-negative (i.e., \( \geq 0 \)). Therefore, the minimum value occurs when \( \left(\cos(\theta) + \frac{1}{2}\right)^2 = 0 \):
\[ 2(0) - \frac{3}{2} = -\frac{3}{2} \]
Thus, the expression \( \cos(2\theta) + 2\cos(\theta) \) is always greater than or equal to \( -\frac{3}{2} \).

### Conclusion
The final result is:
\[ \cos(2\theta) + 2\cos(\theta) \geq -\frac{3}{2} \]