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Question

How do you solve sin2x=sinx ?


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Solution

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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