The complete solution set of the inequality `cos^(-1)(cos 4) gt 3x^(2) - 4x` is

Options

A. `(0, (2 + sqrt(6 pi - 8))/3)`
B. `((2 - sqrt(6 pi - 8))/3 ,0)`
C. `(-2, 2)`
D. `((2 - sqrt(6 pi - 8))/3 ,(2 + sqrt(6 pi - 8))/3)`

Answer

To solve the inequality \( \cos^{-1}(\cos 4) > 3x^2 - 4x \), we will follow these steps:

### Step 1: Determine the value of \( \cos^{-1}(\cos 4) \)

The function \( \cos^{-1}(\cos x) \) has specific properties based on the interval of \( x \):
- If \( x \) is in the interval \( [0, \pi] \), then \( \cos^{-1}(\cos x) = x \).
- If \( x \) is in the interval \( [\pi, 2\pi] \), then \( \cos^{-1}(\cos x) = 2\pi - x \).
- If \( x \) is in the interval \( [2\pi, 3\pi] \), then \( \cos^{-1}(\cos x) = x - 2\pi \).

Since \( 4 \) is in the interval \( [\pi, 2\pi] \) (approximately \( 3.14 < 4 < 6.28 \)), we have:
\[ \cos^{-1}(\cos 4) = 2\pi - 4 \]

### Step 2: Rewrite the inequality

Now we can rewrite the inequality:
\[ 2\pi - 4 > 3x^2 - 4x \]
This can be rearranged to:
\[ 3x^2 - 4x - (2\pi - 4) < 0 \]
which simplifies to:
\[ 3x^2 - 4x + 4 - 2\pi < 0 \]

### Step 3: Find the roots of the quadratic equation

To find the roots of the quadratic equation \( 3x^2 - 4x + (4 - 2\pi) = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 3 \), \( b = -4 \), and \( c = 4 - 2\pi \).

Calculating the discriminant:
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 3 \cdot (4 - 2\pi) = 16 - 12(4 - 2\pi) = 16 - 48 + 24\pi = 24\pi - 32 \]

Now substituting into the quadratic formula:
\[ x = \frac{4 \pm \sqrt{24\pi - 32}}{6} \]
This simplifies to:
\[ x = \frac{2 \pm \sqrt{6\pi - 8}}{3} \]

### Step 4: Determine the intervals

The roots are:
\[ x_1 = \frac{2 - \sqrt{6\pi - 8}}{3}, \quad x_2 = \frac{2 + \sqrt{6\pi - 8}}{3} \]

The quadratic \( 3x^2 - 4x + (4 - 2\pi) \) opens upwards (since the coefficient of \( x^2 \) is positive), so it will be less than zero between the roots:
\[ \frac{2 - \sqrt{6\pi - 8}}{3} < x < \frac{2 + \sqrt{6\pi - 8}}{3} \]

### Final Solution

Thus, the complete solution set of the inequality \( \cos^{-1}(\cos 4) > 3x^2 - 4x \) is:
\[ \left( \frac{2 - \sqrt{6\pi - 8}}{3}, \frac{2 + \sqrt{6\pi - 8}}{3} \right) \]