solve the equation on the interval $0,2\\pi)$. \n$\\cos x - 1 = -\\cos x$\n$x = \\frac{\\pi}{?}…

Explanation:

Step1: Simplify the equation

Add $\cos x$ to both sides of the equation $\cos x - 1=-\cos x$. We get $\cos x+\cos x - 1 = 0$, which simplifies to $2\cos x-1 = 0$. Then, $2\cos x=1$, and $\cos x=\frac{1}{2}$.

Step2: Find the values of (x) in the interval ([0,2\pi))

We know that if $\cos x = a$, then $x = 2k\pi\pm\cos^{-1}(a)$ for $k\in\mathbb{Z}$. Since $\cos x=\frac{1}{2}$ and $\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}$, when (k = 0): (x=\frac{\pi}{3}) (using (x = \cos^{-1}(a))) and (x = 2\pi-\frac{\pi}{3}=\frac{5\pi}{3}) (using (x=2\pi - \cos^{-1}(a)))

Answer:

(x=\frac{\pi}{3},\frac{5\pi}{3})