Solve the inequalities:
(a) `|sin x|=sin x+1`
(b) `|x^(2)-5x| gt |x^(2)|-|5x|`

Answer

Let's solve the inequalities step by step.

### Part (a): Solve the inequality \( | \sin x | = \sin x + 1 \)

1. **Understanding the absolute value**: The absolute value function can be defined in two cases:
- Case 1: \( \sin x \geq 0 \) implies \( | \sin x | = \sin x \)
- Case 2: \( \sin x < 0 \) implies \( | \sin x | = -\sin x \)

2. **Case 1: \( \sin x \geq 0 \)**
- Here, \( | \sin x | = \sin x \)
- The equation becomes \( \sin x = \sin x + 1 \)
- Simplifying gives \( 0 = 1 \), which is a contradiction. Therefore, there are no solutions in this case.

3. **Case 2: \( \sin x < 0 \)**
- Here, \( | \sin x | = -\sin x \)
- The equation becomes \( -\sin x = \sin x + 1 \)
- Rearranging gives \( -2\sin x = 1 \) or \( \sin x = -\frac{1}{2} \)

4. **Finding solutions for \( \sin x = -\frac{1}{2} \)**:
- The general solutions for \( \sin x = -\frac{1}{2} \) are given by:
\[ x = \frac{7\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{11\pi}{6} + 2n\pi, \quad n \in \mathbb{Z} \]

5. **Conclusion for part (a)**:
- The solutions are:
\[ x = \frac{7\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{11\pi}{6} + 2n\pi, \quad n \in \mathbb{Z} \]

### Part (b): Solve the inequality \( |x^2 - 5x| > |x^2| - |5x| \)

1. **Understanding the absolute values**:
- We need to analyze the expression \( |x^2 - 5x| \) and \( |x^2| - |5x| \).
- The absolute value \( |x^2| = x^2 \) for all \( x \).
- The absolute value \( |5x| = 5|x| \).

2. **Case 1: \( x \geq 5 \)**
- Here, \( |x^2 - 5x| = x^2 - 5x \) and \( |5x| = 5x \).
- The inequality becomes:
\[ x^2 - 5x > x^2 - 5x \]
- This is never true, so no solutions in this case.

3. **Case 2: \( 0 \leq x < 5 \)**
- Here, \( |x^2 - 5x| = 5x - x^2 \) (since \( x^2 - 5x < 0 \)).
- The inequality becomes:
\[ 5x - x^2 > x^2 - 5x \]
- Simplifying gives:
\[ 10x - 2x^2 > 0 \implies 2x(5 - x) > 0 \]
- This gives \( 0 < x < 5 \).

4. **Case 3: \( x < 0 \)**
- Here, \( |x^2 - 5x| = x^2 - 5x \) (since both terms are positive).
- The inequality becomes:
\[ x^2 - 5x > x^2 + 5x \]
- Simplifying gives:
\[ -10x > 0 \implies x < 0 \]
- This is always true for \( x < 0 \).

5. **Conclusion for part (b)**:
- The solutions are:
\[ x < 0 \quad \text{or} \quad 0 < x < 5 \]
- In interval notation, the solution is:
\[ (-\infty, 0) \cup (0, 5) \]