Question

solve \\(\cos(x)(\cos(x) - 1) = 0\\).\
\
\\(\bigcirc\\) a. \\(x = \pm \pi n, x = \frac{\pi}{2} \pm 2\pi n\\)\
\\(\bigcirc\\) b. \\(x = \pm \pi n\\)\
\\(\bigcirc\\) c. \\(x = \frac{\pi}{3} \pm 2\pi n, x = \frac{3\pi}{4} \pm 2\pi n\\)\
\\(\bigcirc\\) d. \\(x = \frac{\pi}{2} \pm 2\pi n, x = \frac{3\pi}{2} \pm 2\pi n, x = \pm 2\pi n\\)

Explanation:

Step1: Set each factor to 0

$\cos(x) = 0$ or $\cos(x)-1 = 0$

Step2: Solve $\cos(x)=0$

$x = \frac{\pi}{2} \pm 2\pi n$ or $x = \frac{3\pi}{2} \pm 2\pi n$, which simplifies to $x = \frac{\pi}{2} + \pi n$ (or $x = \pm\frac{\pi}{2} + 2\pi n$, equivalent to $\pm\pi n + \frac{\pi}{2}$; combined, this is $x = \pi n + \frac{\pi}{2}$)

Step3: Solve $\cos(x)-1=0$

$\cos(x)=1$, so $x = 0 \pm 2\pi n = \pm2\pi n$

Step4: Combine all solutions

$x = \frac{\pi}{2} \pm 2\pi n$, $x = \frac{3\pi}{2} \pm 2\pi n$, $x = \pm2\pi n$ (note $\frac{\pi}{2} + \pi n$ covers $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ cases, which matches option D's form)

Answer:

D. $x = \frac{\pi}{2} \pm 2\pi n,x = \frac{3\pi}{2} \pm 2\pi n,x = \pm2\pi n$