if `-1 le x le -1/2 then cos^(-1)(4x^(3)-3x)` equals

Options

A. `3cos^(-1)x`
B. `2pi -3cos^(-1)x`
C. `-2pi +3 cos^(-1)x`
D. none of these

Answer

To solve the problem, we need to evaluate \( \cos^{-1}(4x^3 - 3x) \) given that \( -1 \leq x \leq -\frac{1}{2} \).

### Step-by-step Solution:

1. **Identify the Identity**: We recognize that the expression \( 4x^3 - 3x \) can be related to the cosine of a triple angle. Specifically, we use the identity:
\[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]
This implies that if we let \( x = \cos(\theta) \), then:
\[ 4x^3 - 3x = \cos(3\theta) \]

2. **Set Up the Inverse Cosine**: From the identity, we can write:
\[ \cos^{-1}(4x^3 - 3x) = \cos^{-1}(\cos(3\theta)) \]

3. **Determine the Range of \( \theta \)**: Since \( x \) is in the range \( -1 \leq x \leq -\frac{1}{2} \), we find the corresponding range for \( \theta \):
- If \( x = -1 \), then \( \theta = \pi \).
- If \( x = -\frac{1}{2} \), then \( \theta = \frac{2\pi}{3} \).
Thus, \( \theta \) ranges from \( \frac{2\pi}{3} \) to \( \pi \).

4. **Evaluate the Inverse Cosine**: Since \( \cos^{-1}(\cos(3\theta)) = 3\theta \) when \( 3\theta \) is in the range \( [0, 2\pi] \), we need to check if \( 3\theta \) falls within this range:
- For \( \theta = \frac{2\pi}{3} \), \( 3\theta = 2\pi \).
- For \( \theta = \pi \), \( 3\theta = 3\pi \).
Thus, \( 3\theta \) ranges from \( 2\pi \) to \( 3\pi \).

5. **Adjust for the Range of Inverse Cosine**: Since \( 3\theta \) is outside the range of \( [0, 2\pi] \), we use the periodicity of the cosine function:
\[ \cos^{-1}(\cos(3\theta)) = 3\theta - 2\pi \]
Therefore:
\[ \cos^{-1}(4x^3 - 3x) = 3\theta - 2\pi \]

6. **Express in Terms of \( x \)**: Since \( x = \cos(\theta) \), we have:
\[ \theta = \cos^{-1}(x) \]
Thus:
\[ \cos^{-1}(4x^3 - 3x) = 3\cos^{-1}(x) - 2\pi \]

### Final Answer:
\[ \cos^{-1}(4x^3 - 3x) = 3\cos^{-1}(x) - 2\pi \]