To solve the equation [tex]\(\cos(x) (\cos(x) - 1) = 0\)[/tex], we need to find the values of [tex]\(x\)[/tex] for which the equation holds true.
Let's break this down step-by-step:
1. Identify when the equation equals zero:
The product of two terms equals zero if and only if at least one of the terms is zero. So, we have two cases:
[tex]\[
\cos(x) = 0 \quad \text{or} \quad \cos(x) - 1 = 0
\][/tex]
2. Solve [tex]\(\cos(x) = 0\)[/tex]:
[tex]\[
\cos(x) = 0
\][/tex]
Cosine is zero at odd multiples of [tex]\(\frac{\pi}{2}\)[/tex]:
[tex]\[
x = \frac{\pi}{2} + k\pi \quad \text{for integer} \; k
\][/tex]
This can be written as:
[tex]\[
x = \frac{\pi}{2} + 2n\pi \quad \text{or} \quad x = \frac{3\pi}{2} + 2n\pi \quad \text{for integer} \; n
\][/tex]
3. Solve [tex]\(\cos(x) - 1 = 0\)[/tex]:
[tex]\[
\cos(x) - 1 = 0 \implies \cos(x) = 1
\][/tex]
Cosine is 1 at even multiples of [tex]\(\pi\)[/tex]:
[tex]\[
x = 2n\pi \quad \text{for integer} \; n
\][/tex]
Combining the solutions we obtained:
- From [tex]\(\cos(x) = 0\)[/tex], we get:
[tex]\[
x = \frac{\pi}{2} + 2n\pi \quad \text{and} \quad x = \frac{3\pi}{2} + 2n\pi \quad \text{for integer} \; n
\][/tex]
- From [tex]\(\cos(x) = 1\)[/tex], we get:
[tex]\[
x = 2n\pi \quad \text{for integer} \; n
\][/tex]
Now, we will combine these two sets of solutions:
[tex]\[
x = \frac{\pi}{2} + 2n\pi, \quad x = \frac{3\pi}{2} + 2n\pi, \quad x = 2n\pi \quad \text{for integer} \; n
\][/tex]
To match these general forms with the provided options, let's recheck each one:
- [tex]\(\textbf{Option A}\)[/tex] mentions:
[tex]\[
x = \pm \pi n, \quad x = \frac{\pi}{2} \pm 2\pi n
\][/tex]
This is incorrect because it does not include [tex]\(\frac{3\pi}{2} + 2n\pi\)[/tex].
- [tex]\(\textbf{Option B}\)[/tex] mentions:
[tex]\[
x = \pm \pi n
\][/tex]
This is incorrect as it misses [tex]\(\frac{\pi}{2}\)[/tex] and [tex]\(\frac{3\pi}{2}\)[/tex] solutions.
- [tex]\(\textbf{Option C}\)[/tex] mentions:
[tex]\[
x = \frac{\pi}{2} \pm 2\pi n, \quad x = \frac{3\pi}{2} \pm 2\pi n, \quad x = \pm 2\pi n
\][/tex]
This correctly includes all our solutions since:
[tex]\[
x = \frac{\pi}{2} + 2n\pi, \;\; x = \frac{3\pi}{2} + 2n\pi, \;\; x = 2n\pi
\][/tex]
- [tex]\(\textbf{Option D}\)[/tex] mentions:
[tex]\[
x = \frac{\pi}{3} \pm 2\pi n, \quad x = \frac{3\pi}{4} \pm 2\pi n
\][/tex]
This is incorrect as these solutions do not correspond to the zeros of [tex]\(\cos(x)(\cos(x) - 1)\)[/tex].
Thus, the correct solution is:
[tex]\(\boxed{\text{C}}\)[/tex]
