solve for exact solutions over the interval $0,2\\pi)$. \n$\\cos 2x=\\frac{1}{2}$ \nselect the correct…

solve for exact solutions over the interval $0,2\\pi)$. \n$\\cos 2x=\\frac{1}{2}$ \nselect the correct choice below and, if necessary, fill in the answer box to complete your choice. \na. the solution set is (simplify your answer. type an exact answer, using $\\pi$ as needed. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed ) \nb. the solution set is $\\varnothing$.

solve for exact solutions over the interval $0,2\\pi)$. \n$\\cos 2x=\\frac{1}{2}$ \nselect the correct choice below and, if necessary, fill in the answer box to complete your choice. \na. the solution set is (simplify your answer. type an exact answer, using $\\pi$ as needed. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed ) \nb. the solution set is $\\varnothing$.

Answer

Explanation:

Step1: Find general solutions for (2x)

We know that if (\cos\theta=\frac{1}{2}), then (\theta = 2k\pi\pm\frac{\pi}{3}), (k\in\mathbb{Z}). Here (\theta = 2x), so (2x=2k\pi\pm\frac{\pi}{3}).

Step2: Solve for (x)

Divide each equation by (2): (x = k\pi\pm\frac{\pi}{6}).

Step3: Find solutions in the interval ([0,2\pi))

When (k = 0): (x=\frac{\pi}{6}) (from (x=k\pi+\frac{\pi}{6})) and (x =-\frac{\pi}{6}+2\pi=\frac{11\pi}{6}) (from (x=k\pi - \frac{\pi}{6}), since (x\geq0), we use (x=(k + 2)\pi-\frac{\pi}{6}) with (k=- 1) equivalent to (x = 2\pi-\frac{\pi}{6})). When (k = 1): (x=\pi+\frac{\pi}{6}=\frac{7\pi}{6}) (from (x=k\pi+\frac{\pi}{6})) and (x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}) (from (x=k\pi-\frac{\pi}{6})).

Answer:

A. The solution set is (\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6})