Y=|sin(3x-П:2)-1|

Y = | sin(3x - π/2) - 1 |

Simplify:

• Use sin(α − π/2) = −cos α, so sin(3x − π/2) = −cos(3x).
• Then Y = | −cos(3x) − 1 | = | cos(3x) + 1 |.

Since cos(3x) ∈ [−1, 1], cos(3x) + 1 ∈ [0, 2], which is already nonnegative. Thus the absolute value is redundant:

Y = cos(3x) + 1.

Key properties:

• Domain: all real x.
• Range: [0, 2].
• Period: 2π/3 (period of cos(3x) is 2π/3).
• Zeros: when cos(3x) = −1 → 3x = π + 2πk → x = π/3 + 2πk/3, k ∈ Z.
• Maximum: 2 at cos(3x) = 1 → x = 2πk/3.
• Minimum: 0 at cos(3x) = −1 → x = π/3 + 2πk/3.

If you want, I can sketch the graph or provide example x-values.