Solve [tex]\sin(x)(\sin(x)-1)=0[/tex]

A. [tex]x= \pm \pi n[/tex]

B. [tex]x=\frac{\pi}{3} \pm 2 \pi n, \quad x=\frac{3 \pi}{4} \pm 2 \pi n[/tex]

C. [tex]x=\frac{\pi}{2} \pm 2 \pi n, \quad x=\frac{3 \pi}{2} \pm 2 \pi n[/tex]

D. [tex]x= \pm \pi n, \quad x=\frac{\pi}{2} \pm 2 \pi n[/tex]



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]