`cos 2x + 2cos x` 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 expression \( \cos 2x + 2\cos x \) and determine its range, we can follow these steps:

### Step 1: Use the double angle formula
We know that:
\[ \cos 2x = 2\cos^2 x - 1 \]
Substituting this into the expression gives:
\[ \cos 2x + 2\cos x = (2\cos^2 x - 1) + 2\cos x \]

### Step 2: Simplify the expression
Now, we can simplify the expression:
\[ = 2\cos^2 x + 2\cos x - 1 \]
This can be rearranged as:
\[ = 2\cos^2 x + 2\cos x - 1 \]

### Step 3: Complete the square
Next, we can complete the square for the quadratic in terms of \( \cos x \):
\[ = 2\left(\cos^2 x + \cos x\right) - 1 \]
To complete the square inside the parentheses:
\[ \cos^2 x + \cos x = \left(\cos x + \frac{1}{2}\right)^2 - \frac{1}{4} \]
Substituting this back in gives:
\[ = 2\left(\left(\cos x + \frac{1}{2}\right)^2 - \frac{1}{4}\right) - 1 \]
\[ = 2\left(\cos x + \frac{1}{2}\right)^2 - \frac{1}{2} - 1 \]
\[ = 2\left(\cos x + \frac{1}{2}\right)^2 - \frac{3}{2} \]

### Step 4: Determine the range
The term \( \left(\cos x + \frac{1}{2}\right)^2 \) is always non-negative (i.e., \( \geq 0 \)). Therefore, the minimum value of \( 2\left(\cos x + \frac{1}{2}\right)^2 \) is 0. Thus, the minimum value of the entire expression is:
\[ 0 - \frac{3}{2} = -\frac{3}{2} \]
The maximum value occurs when \( \left(\cos x + \frac{1}{2}\right)^2 \) is maximized, which can go up to \( 2 \) (when \( \cos x = 1 \)). So, the maximum value of the expression is:
\[ 2(2) - \frac{3}{2} = 4 - \frac{3}{2} = \frac{8}{2} - \frac{3}{2} = \frac{5}{2} \]

### Conclusion
Thus, the expression \( \cos 2x + 2\cos x \) has a minimum value of \( -\frac{3}{2} \) and a maximum value of \( \frac{5}{2} \). Therefore, the expression is always greater than or equal to \( -\frac{3}{2} \).

### Final Answer
The correct option is that \( \cos 2x + 2\cos x \) is always greater than or equal to \( -\frac{3}{2} \).

---