Trigonometry (11th Edition) Clone

The solution set is $$\{\frac{\pi}{3}+2\pi n, \frac{2\pi}{3}+2\pi n, \frac{4\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n, n\in Z\}$$
$$4\cos^2 x-1=0$$ 1) Solve the equation over the interval $[0,2\pi)$ $$4\cos^2 x-1=0$$ $$\cos^2x=\frac{1}{4}$$ $$\cos x=\pm\frac{1}{2}$$ - For $\cos x=\frac{1}{2}$ Over the interval $[0,2\pi)$, there are two values of $x$ where $\cos x=\frac{1}{2}$, which are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$. - For $\cos x=-\frac{1}{2}$ Over the interval $[0,2\pi)$, there are two values of $x$ where $\cos x=-\frac{1}{2}$, which are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$. Therefore, overall, $$x=\{\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the cosine function is $2\pi$. - We apply it to each solution found in step 1. The results are the solution set. So the solution set is $$\{\frac{\pi}{3}+2\pi n, \frac{2\pi}{3}+2\pi n, \frac{4\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n, n\in Z\}$$