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[rapport1]

Or if it can't be found, could someone pls solve it?

 

[KloppsAndRobbers]

View attachment 43266

Or if it can't be found, could someone pls solve it?

sin(3θ) = 2sinθ
sin(2θ + θ) = 2sinθ
sin2θcosθ + cos2θsinθ = 2sinθ
(sinθcosθ + cosθsinθ)cosθ + sinθ(1-2sin^2θ) = 2sinθ
sinθcos^2θ + sinθcos^2θ + sinθ - 2sin^3θ = 2sinθ
sinθ(1-sin^2θ) + sinθ(1-sin^2θ) + sinθ - 2sin^3θ = 2sinθ
sinθ - sin^3θ + sinθ - sin^3θ + sinθ - 2sin^3θ = 2sinθ
3sinθ - 4sin^3θ = 2sinθ
4sin^3θ - sinθ = 0
sinθ(4sin^2θ - 1) = 0

sinθ = 0.
θ = 0, π, 2π.

4sin^2θ = 1
sin^2θ = 1/4
sinθ = ±1/2
θ = π/6, 5π/6, 7π/6, 11π/6

Therefore, θ = 0, π/6, 5π/6, π, 7π/6, 11π/6, 2π.