Answer :
To solve the equation [tex]\(\sin(x)(\sin(x) - 1) = 0\)[/tex], we need to set each factor in the product to zero and solve for [tex]\(x\)[/tex].
First, we consider the equation:
[tex]\[ \sin(x) = 0 \][/tex]
The sine function is zero at integer multiples of [tex]\(\pi\)[/tex]:
[tex]\[ x = n\pi \quad \text{where } n \in \mathbb{Z} \][/tex]
Next, we solve the second factor:
[tex]\[ \sin(x) - 1 = 0 \implies \sin(x) = 1 \][/tex]
The sine function equals 1 at:
[tex]\[ x = \frac{\pi}{2} + 2k\pi \quad \text{where } k \in \mathbb{Z} \][/tex]
Now let's summarize the solutions we have found:
1. From [tex]\(\sin(x) = 0\)[/tex]: [tex]\(x = n\pi \)[/tex] where [tex]\( n \)[/tex] is an integer.
2. From [tex]\(\sin(x) = 1\)[/tex]: [tex]\( x = \frac{\pi}{2} + 2k\pi \)[/tex] where [tex]\( k \)[/tex] is an integer.
These solutions together cover all the values of [tex]\(x\)[/tex] that satisfy the original equation:
[tex]\[ x = n\pi, \text{ and } x = \frac{\pi}{2} + 2k\pi \quad \text{where } n, k \in \mathbb{Z} \][/tex]
Expressing the solution in a more convenient form, we can identify the correct answer among the provided options. The full set of solutions can be written compactly as:
[tex]\[ x = \pm n\pi, n \in \mathbb{Z} \quad \text{and} \quad x = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} \][/tex]
Considering these formulations and the provided options, the correct answer is:
D. [tex]\(x= \pm \pi n, x=\frac{\pi}{2} \pm 2 \pi n\)[/tex]
First, we consider the equation:
[tex]\[ \sin(x) = 0 \][/tex]
The sine function is zero at integer multiples of [tex]\(\pi\)[/tex]:
[tex]\[ x = n\pi \quad \text{where } n \in \mathbb{Z} \][/tex]
Next, we solve the second factor:
[tex]\[ \sin(x) - 1 = 0 \implies \sin(x) = 1 \][/tex]
The sine function equals 1 at:
[tex]\[ x = \frac{\pi}{2} + 2k\pi \quad \text{where } k \in \mathbb{Z} \][/tex]
Now let's summarize the solutions we have found:
1. From [tex]\(\sin(x) = 0\)[/tex]: [tex]\(x = n\pi \)[/tex] where [tex]\( n \)[/tex] is an integer.
2. From [tex]\(\sin(x) = 1\)[/tex]: [tex]\( x = \frac{\pi}{2} + 2k\pi \)[/tex] where [tex]\( k \)[/tex] is an integer.
These solutions together cover all the values of [tex]\(x\)[/tex] that satisfy the original equation:
[tex]\[ x = n\pi, \text{ and } x = \frac{\pi}{2} + 2k\pi \quad \text{where } n, k \in \mathbb{Z} \][/tex]
Expressing the solution in a more convenient form, we can identify the correct answer among the provided options. The full set of solutions can be written compactly as:
[tex]\[ x = \pm n\pi, n \in \mathbb{Z} \quad \text{and} \quad x = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} \][/tex]
Considering these formulations and the provided options, the correct answer is:
D. [tex]\(x= \pm \pi n, x=\frac{\pi}{2} \pm 2 \pi n\)[/tex]
