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expeople1 [14]
2 years ago
6

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

Mathematics
1 answer:
Nastasia [14]2 years ago
3 0

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} .

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