Problem 35 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

The solutions of the equation \(\sin x(\sin x+1)=0\) are \(x = n\pi\) and \(x = (2n+1)\pi + 3\pi/2\), for all integers n.

Step 1: Set each factor equal to zero

The equation \(\sin x(\sin x+1)=0\) can be split into two separate equations, using the principle that a product of factors is zero if and only if at least one of the factors is zero. Thus, the two equations will be \(\sin x = 0\) and \(\sin x + 1 = 0\).

Step 2: Solve for x in the first equation

In our first equation, \(\sin x = 0\), the values of x that satisfy this condition are \(x = n\pi\), where n is an integer. This is because the sine function oscillates between -1 and 1 and is equal to zero at every multiple of \(\pi\).

Step 3: Solve for x in the second equation

For the second equation, check the values of x where the sine function equals -1. Subtract 1 from both sides of the equation \(\sin x + 1 = 0\) to get \(\sin x = -1\). The sine function equals -1 at \(x = (2n+1)\pi + 3\pi/2\), where n is any integer.

Step 4: Combine all solutions

After solving both equations, combine all the values of x from steps 2 and 3 to obtain the complete solution set. The solutions are \(x = n\pi\) and \(x = (2n+1)\pi + 3\pi/2\), for all integers n.