The true solution set of the inequality, `sqrt(5x-6-x^2)+(pi/2int_0^x dz)>x int_0^pisin^2xdx` is:

Options

A. `x in(2,3)`
B. `x in(-oo,2)uu(3,oo)`
C. `x in((5)/(2),3)`
D. none of these

Answer

To solve the inequality

\[ \sqrt{5x - 6 - x^2} + \frac{\pi}{2} \int_0^x dz > x \int_0^\pi \sin^2 x \, dx, \]

we'll break it down step by step.

### Step 1: Simplify the Left Side
We start with the left side of the inequality:

\[ \sqrt{5x - 6 - x^2} + \frac{\pi}{2} \int_0^x dz. \]

The integral \(\int_0^x dz\) evaluates to \(x\), so we can rewrite the left side as:

\[ \sqrt{5x - 6 - x^2} + \frac{\pi}{2} x. \]

### Step 2: Simplify the Right Side
Now, let's simplify the right side:

\[ x \int_0^\pi \sin^2 x \, dx. \]

Using the identity \(\sin^2 x = \frac{1 - \cos(2x)}{2}\), we can compute the integral:

\[ \int_0^\pi \sin^2 x \, dx = \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_0^\pi = \frac{1}{2} \left[ \pi - 0 \right] = \frac{\pi}{2}. \]

Thus, the right side becomes:

\[ x \cdot \frac{\pi}{2} = \frac{\pi}{2} x. \]

### Step 3: Set Up the Inequality
Now we can rewrite the original inequality:

\[ \sqrt{5x - 6 - x^2} + \frac{\pi}{2} x > \frac{\pi}{2} x. \]

This simplifies to:

\[ \sqrt{5x - 6 - x^2} > 0. \]

### Step 4: Solve the Square Root Inequality
The expression under the square root must be non-negative:

\[ 5x - 6 - x^2 \geq 0. \]

Rearranging gives:

\[ -x^2 + 5x - 6 \geq 0. \]

Factoring the quadratic:

\[ -(x^2 - 5x + 6) \geq 0 \implies -(x - 2)(x - 3) \geq 0. \]

This implies:

\[ (x - 2)(x - 3) \leq 0. \]

### Step 5: Find the Solution Set
The solution to the inequality \((x - 2)(x - 3) \leq 0\) is found by testing intervals:

1. For \(x < 2\), both factors are negative, so the product is positive.
2. For \(2 \leq x \leq 3\), the product is non-positive.
3. For \(x > 3\), both factors are positive, so the product is positive.

Thus, the solution set is:

\[ x \in [2, 3]. \]

### Final Answer
The true solution set of the inequality is:

\[ \boxed{[2, 3]}. \]