If `x in [-1, 0)` then find the value of `cos^(-1) (2x^(2) -1) -2 sin^(-1) x`

Explanation

To solve the expression \( \cos^{-1}(2x^2 - 1) - 2 \sin^{-1}(x) \) for \( x \in [-1, 0) \), we can follow these steps:

### Step 1: Simplify \( \cos^{-1}(2x^2 - 1) \)

Using the identity:
\[ \cos^{-1}(2x^2 - 1) = 2 \cos^{-1}(x) \]
This identity holds because \( 2x^2 - 1 = \cos(2\theta) \) where \( \theta = \cos^{-1}(x) \).

### Step 2: Substitute into the expression

Now we can substitute this back into the original expression:
\[ \cos^{-1}(2x^2 - 1) - 2 \sin^{-1}(x) = 2 \cos^{-1}(x) - 2 \sin^{-1}(x) \]

### Step 3: Factor out the common term

We can factor out the 2:
\[ = 2 (\cos^{-1}(x) - \sin^{-1}(x)) \]

### Step 4: Use the identity \( \cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2} \)

From the identity, we know:
\[ \cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x) \]
Thus, we can substitute:
\[ \cos^{-1}(x) - \sin^{-1}(x) = \left(\frac{\pi}{2} - \sin^{-1}(x)\right) - \sin^{-1}(x) = \frac{\pi}{2} - 2\sin^{-1}(x) \]

### Step 5: Substitute back into the expression

Now substituting this back, we have:
\[ 2 \left(\frac{\pi}{2} - 2\sin^{-1}(x)\right) = \pi - 4\sin^{-1}(x) \]

### Step 6: Evaluate for \( x \in [-1, 0) \)

Since \( x \) is in the range of \([-1, 0)\), \( \sin^{-1}(x) \) will yield values in \([-\frac{\pi}{2}, 0)\). Thus, we can conclude that:
\[ \cos^{-1}(2x^2 - 1) - 2 \sin^{-1}(x) = \pi - 4\sin^{-1}(x) \]

### Final Result

The final expression simplifies to:
\[ \pi - 4\sin^{-1}(x) \]