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Chapter 5: Problem 23

Solve the equation. $$\sin x(\sin x+1)=0$$

Short Answer

Expert verified
The solutions for the equation are \( x = n\pi \) and \( x = (2n+1)\pi \), where \( n \) is an integer.

Step by step solution

01

Factor the Equation

The equation can be factored into \( \sin x \) and \( \sin x+1 \). Doing so gives us two equations: \( \sin x = 0 \) and \( \sin x + 1 = 0 \).
02

Solve the First Equation

For the first equation \( \sin x = 0 \), the solutions are given by solving \( x = \arcsin 0 \). Thus, the solution to the equation \( \sin x = 0 \) is \( x = n\pi \), where \( n \) is an integer.
03

Solve the Second Equation

For the second equation \( \sin x + 1 = 0 \), the solutions are given by solving \( x = \arcsin (-1) \). Thus, the solution to the equation \( \sin x + 1 = 0 \) is \( x = (2n+1)\pi \), where \( n \) is an integer.

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