If `cos^(-1)(cos 4)gt3x^(2)-4x` then

Options

A. `x in (-oo,(2-sqrt(6pi-8))/(3)`
B. `x in ((2+sqrt(6pi-87))/(3),oo)`
C. `x in ((2+sqrt(6pi-8))/(3),oo)`
D. `x in ((2-sqrt(p6pi-8))/(3),(2+sqrt(6pi-8))/(3)`

Explanation

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

### Step 1: Simplify \( \cos^{-1}(\cos 4) \)

The function \( \cos^{-1}(\cos \theta) \) returns \( \theta \) if \( \theta \) is in the range \( [0, \pi] \). Since \( 4 \) is not in this range, we need to find an equivalent angle within \( [0, \pi] \).

The angle \( 4 \) can be expressed as:
\[ 4 = 2\pi - (2\pi - 4) \]
Since \( 2\pi - 4 \) is in the range \( [0, \pi] \), we have:
\[ \cos^{-1}(\cos 4) = 2\pi - 4 \]

### Step 2: Rewrite the inequality

Now we can rewrite the original inequality:
\[ 2\pi - 4 > 3x^2 - 4x \]

### Step 3: Rearrange the inequality

Rearranging gives us:
\[ 3x^2 - 4x < 2\pi - 4 \]

### Step 4: Set up the quadratic inequality

We can express this as:
\[ 3x^2 - 4x - (2\pi - 4) < 0 \]

### Step 5: Solve the quadratic equation

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

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

Now substituting back into the quadratic formula:
\[ x = \frac{4 \pm \sqrt{24\pi - 32}}{6} \]

### Step 6: Find the critical points

The critical points are:
\[ x_1 = \frac{4 + \sqrt{24\pi - 32}}{6}, \quad x_2 = \frac{4 - \sqrt{24\pi - 32}}{6} \]

### Step 7: Determine the intervals

The quadratic \( 3x^2 - 4x - (2\pi - 4) \) opens upwards (since \( a > 0 \)), so it will be less than zero between its roots \( x_2 \) and \( x_1 \).

### Final Result

Thus, the solution to the inequality \( \cos^{-1}(\cos 4) > 3x^2 - 4x \) is:
\[ x_2 < x < x_1 \]