Question

Find all solutions to the equation in the interval [0,2pi]. Enter the solutions in increasing order. sin 2x = sin x

please explain how you got the answer

Answer 1

Explanation:

`sin2x=sinx\\sin2x-sinx=0\\2*sinx*cosx-sinx=0\\sinx*(2*cosx-1)-0\\sinx=0 \\x=\pi n\ \ \ \ \ x\in[0;2\pi ]\ \ \ \ \Rightarrow\\x_1=0\ \ \ \ x_2=\pi .\\2*cosx-1=0\\2*cosx=1\ |:2\\cosx=(1)/(2) \\x=б(\pi )/(3) +2\pi n\ \ \ \ \ x\in[0;2\pi ]\ \ \ \ \Rightarrow\\x_3=(\pi )/(3) \ \ \ \ \ x_4=(5\pi )/(3).\\`

`Answer: x=0,\ (\pi )/(3) ,\ 1\pi ,\ (5\pi )/(3) .`