If sinθ=sin2θ then find the principal solution of θ

Steps to solve If sinθ=sin2θ then find the principal solution of θ

sinθ=sin2θ

⇒θ=nπ+(-1)^n(2θ)

⇒θ-(-1)^n(2θ)=nπ

⇒θ{1-(-1)^n(2)}=nπ

⇒θ=nπ/1-(-1)^n(2)

Put n=0
⇒θ=0.π/1-(-1)^0(2)=0

Put n=1
⇒θ=1.π/1-(-1)^1(2)=π/3

Put n=2
⇒θ=2.π/1-(-1)^2(2)=-2π

Put n=3
⇒θ=3.π/1-(-1)^3(2)=3π

Put n=5
⇒θ=5.π/1-(-1)^5(2)=5π/3

Hence the principal solution for sinθ=sin2θ is 0, π/3, -2π, 3π, 5π/3

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