How do you solve sin⁡2x=sin⁡x?

Question

Philip Williams · Accepted Answer

sin⁡2x=sin⁡x  2sin⁡xcos⁡x=sin⁡x  2sin⁡xcos⁡x−sin⁡x=0  sin⁡x(2cos⁡x−1)=0  Solution A: sin⁡x=0⇒x=kπ,k∈Z  Solution B: 2cos⁡x=1⇒cos⁡x=12,x=±π3+2kπ=π3+2kπ=π3(6k±1), k∈Z  ∴x=kπ or x=π3(6k±1),k∈Z

Paineow · Accepted Answer

Factor sin⁡(x) out of 2sin⁡(x)cos⁡(x)−sin⁡(x)  sin⁡(x)(2cos⁡(x)−1)=0  If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.  sin⁡(x)=0  2cos⁡(x)−1=0  Set the first factor equal to 0 and solve.  x=2πn,π+2πn for any integer n  Set the next factor equal to 0 and solve.  x=π3+2πn, for any integer n  The final solution is all the values that make sin⁡(x)(2cos⁡(x)−1)=0 true.  x=2πn,π+2πn,π3+2πn,5π3+2πn, for any integer n  The finall solution is all the values that make sin⁡(x)(2cos⁡(x)−1)=0 true.  x=2πn,π+2πn,π3+2πn,5π3+2πn, for any integer n  Consolidate 2πn and π+2πn to πn  x=πn,π3+2πn,5π3+2πn for any integer n

user_27qwe · Accepted Answer

Very good answer

How do you solve \sin2x=\sin x?

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