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Solve 2sin^2(2x)-sin2x = 0 over [0,2π[

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2sin^2(2x)-sin2x = 0

sin2x(2sin2x-1) = 0

sin2x = 0

2x = 0 or π

2sin2x-1 = 0

2sin2x = 1

sin2x = 1/2

2x = π/6 or 5π/6

Dividing the above values by 2 to get x = {0, π/2, π/12, 5π/12} over [0,2π[

However, the actual solution to the equation is {0, π/2, π/12, 3π/12, 5π/12, π, 13π/12, 17π/12}. Why didn't I also get 3π/12, π, 13π/12, and 17π/12 and what did I do wrong?

You're pretty close. First of all, given that the argument of the sin is θ =2x, then I don't think x = 3𝛑/12 should be included since this would give sin (3𝛑/6) = sin(𝛑/2) ≠ 0.

For the rest of the roots, as you found, there are 4 roots for θ in [0, 2𝛑], namely

for sin(θ) = 0, roots are 0 and 𝛑

and for sin(θ) = 1/2, roots are 𝛑/6 and 5𝛑/6.

Now, for each of these roots, θ_r, we know that θ_r + 2𝛑 is also a root, since we have just completed exactly 1 circuit of the unit circle, so that for

θ = 0 --> 2𝛑, 4𝛑, 6𝛑, etc. are also roots

θ = 𝛑 --> 3𝛑, 5𝛑, 7𝛑, etc.

θ = 𝛑/6 --> 13𝛑/6, 25𝛑/6, etc. where 2𝛑 = 12𝛑/6 had been added

θ = 5𝛑/6 --> 17𝛑/6, 27𝛑/6, etc.

Dividing the above roots by 2 to get x, and confining the values to [0, 2𝛑], gives the answer you want.

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