(6) if \( (x+1, y-2)=(3,1) \), find the values of \( x \) and \( y \) ii) Find the domain and range of the function \( \sin x \)

To solve the first part of your question: Given the equation \( (x+1, y-2) = (3, 1) \), we can equate the components: 1. \( x + 1 = 3 \) 2. \( y - 2 = 1 \) Now, we can solve for \( x \) and \( y \): 1. For \( x \): \[ x + 1 = 3 \implies x = 3 - 1 = 2 \] 2. For \( y \): \[ y - 2 = 1 \implies y = 1 + 2 = 3 \] Thus, the values are: \[ x = 2 \quad \text{and} \quad y = 3 \] --- For the second part of your question regarding the function \( \sin x \): **Domain of \( \sin x \)**: The sine function is defined for all real numbers. Therefore, the domain of \( \sin x \) is: \[ \text{Domain: } (-\infty, \infty) \] **Range of \( \sin x \)**: The sine function oscillates between -1 and 1. Therefore, the range of \( \sin x \) is: \[ \text{Range: } [-1, 1] \] In summary: - Values of \( x \) and \( y \): \( x = 2 \), \( y = 3 \) - Domain of \( \sin x \): \( (-\infty, \infty) \) - Range of \( \sin x \): \( [-1, 1] \)