Question

(Sinx-1/2)(Sinx+1)=0

Answer 1

$(sinx- \frac{1}{2} )(sinx+1)=0 \\sinx- \frac{1}{2}=0 \\sinx= \frac{1}{2} \\x_1= \frac{\pi}{6} +2\pi n,\ n \in Z \\x_2= \frac{5\pi}{6} +2\pi n,\ n \in Z \\sinx+1=0 \\sinx=-1 \\x_3=- \frac{\pi}{2} +2\pi n,\ n \in Z$

Answer 2

To solve the equation $(\sin(x) - \frac{1}{2})(\sin(x) + 1) = 0$, you can use the zero-product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be equal to zero. So, you can set each factor equal to zero and solve for (x:

1. $\sin(x) - \frac{1}{2} = 0$:

   Add $1/2$ to both sides to isolate $\sin(x)$:

   $\sin(x) = \frac{1}{2}$

   Now, find the values of $x$ where $\sin(x) = \frac{1}{2}$. The sine function is equal to $1/2$ at certain angles. The common values are $\frac{\pi}{6} + 2\pi n$ and $\frac{5\pi}{6} + 2\pi n$, where $n$ is an integer.

2. $\sin(x) + 1 = 0$:

   Subtract 1 from both sides to isolate $\sin(x)$:

   $\sin(x) = -1$

   Now, find the values of $x$ where $\sin(x) = -1$. The sine function is equal to $-1$ at $\frac{3\pi}{2} + 2\pi n$, where $n$ is an integer.

So, the solutions to the equation $(\sin(x) - \frac{1/2})(\sin(x) + 1) = 0$ are:

1. $x = \frac{\pi}{6} + 2\pi n$
2. $x = \frac{5\pi}{6} + 2\pi n$
3. $x = \frac{3\pi}{2} + 2\pi n$

Where $n$ is an integer, and it can take on any integer value to generate different solutions.