How do you solve sin2x=sinx ?
Compute the required value.
Given: sin2x=sinx
Since, sin2x=2sinxcosx
So, 2sinxcosx=sinx
⇒2sinxcosx-sinx=0
⇒ sinx2cosx-1=0
Thus, sinx=0 or 2cosx-1=0
If sinx=0,
Since sinθ is 0, for θ=π and sinx is periodic in 2π period.
then x=kπ,k=0,±1,±2.......
If 2cosx-1=0
⇒ cosx=12
Since cosθ is 12, for θ=π3 and cosx is periodic in 2π period.
then, x=π3+2kπ,k=0,±1,±2........
Hence the solutions in the interval 0,2π are: 0,π,2π,π3,5π3
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