Question

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

Answer 1

Step-by-step explanation:

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

$Answer: x=0,\ \frac{\pi }{3} ,\ 1\pi ,\ \frac{5\pi }{3} .$