Trigonometry (11th Edition) Clone

The solution set is $$\{0,\frac{2\pi}{3},\frac{4\pi}{3}\}$$
$$\cos2x-\cos x=0$$ over interval $[0,2\pi)$ 1) In this case, only the interval for $x$, which is $[0,2\pi)$, is necessary, as you will see in step 2 that $\cos2x$ would be changed to a function of $x$ only. $$x\in[0,2\pi)$$ 2) Now consider back the equation $$\cos2x-\cos x=0$$ Here we see that $\cos x$ is a trigonometric function of $x$, but $\cos2x$ is that of $2x$. Thus it is essential to change $\cos2x$ to a trigonometric function of $x$ by using the identity $\cos2x=2\cos^2 x-1$ $$2\cos^2x-1-\cos x=0$$ $$(2\cos^2x-2\cos x)+(\cos x-1)=0$$ $$(2\cos x+1)(\cos x-1)=0$$ $$\cos x=-\frac{1}{2}\hspace{1cm}\text{or}\hspace{1cm}\cos x=1$$ For $\cos x=1$, over the interval $[0,2\pi)$, there are 1 value whose $\cos$ equals $1$, which is $\{0\}$ For $\cos x=-\frac{1}{2}$, over the interval $[0,2\pi)$, there are 2 values whose $\cos$ equals $-\frac{1}{2}$, which are $\{\frac{2\pi}{3},\frac{4\pi}{3}\}$ Combining the solutions of 2 cases where $\cos x=1$ or $\cos x=-\frac{1}{2}$, we end up with the solution set: $$x=\{0,\frac{2\pi}{3},\frac{4\pi}{3}\}$$