solve the equation for the interval $0,2\\pi)$. $2\\sin^{2}(x)=\\sin(x)$ o a. $\\{\\frac{\\pi}{2},\\frac{3\\p…

solve the equation for the interval $0,2\\pi)$. $2\\sin^{2}(x)=\\sin(x)$ o a. $\\{\\frac{\\pi}{2},\\frac{3\\pi}{2},\\frac{\\pi}{3},\\frac{2\\pi}{3}\\}$ o b. $\\{\\frac{\\pi}{3},\\frac{2\\pi}{3}\\}$ o c. $\\{\\frac{\\pi}{6},\\frac{5\\pi}{6}\\}$ o d. $\\{0,\\pi,\\frac{\\pi}{6},\\frac{5\\pi}{6}\\}$

solve the equation for the interval $0,2\\pi)$. $2\\sin^{2}(x)=\\sin(x)$ o a. $\\{\\frac{\\pi}{2},\\frac{3\\pi}{2},\\frac{\\pi}{3},\\frac{2\\pi}{3}\\}$ o b. $\\{\\frac{\\pi}{3},\\frac{2\\pi}{3}\\}$ o c. $\\{\\frac{\\pi}{6},\\frac{5\\pi}{6}\\}$ o d. $\\{0,\\pi,\\frac{\\pi}{6},\\frac{5\\pi}{6}\\}$

Answer

Explanation:

Step1: Rearrange the equation

Move all terms to one side: (2\sin^{2}(x)-\sin(x) = 0). Factor out (\sin(x)): (\sin(x)(2\sin(x)-1)=0).

Step2: Solve (\sin(x)=0)

For (\sin(x)=0) and (x\in[0,2\pi)), using the unit - circle, (x = 0) or (x=\pi).

Step3: Solve (2\sin(x)-1 = 0)

First, solve (2\sin(x)-1 = 0) for (\sin(x)): (\sin(x)=\frac{1}{2}). Using the unit - circle, for (\sin(x)=\frac{1}{2}) and (x\in[0,2\pi)), (x=\frac{\pi}{6}) or (x = \frac{5\pi}{6}).

Answer:

D. (\left{0,\pi,\frac{\pi}{6},\frac{5\pi}{6}\right})