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Mathematics 17 Online
OpenStudy (anonymous):

Find all solutions in the interval [0, 2π). 2 sin2x = sin x Answers: x = π/3, 2π/3 x = π/2, 3π/2, π/3, 2π/3 x = 0, π, π/6, 5π/6 x= π/6, 5π/6 I want to say it is the final one. I know π/6 is an answer I just don't know how to find others.

OpenStudy (anonymous):

2sin(2x)=sinx =>2(2sinx cosx)=sinx =>sinx(4cosx-1)=0 either sinx=0 or cosx=1/4, both satisfies the above equation.

OpenStudy (anonymous):

Sorry I forgot the power sign on the sin. It is 2 sin^2x = sin x

OpenStudy (anonymous):

ohk.. so, sinx(2sinx-1)=0 either sinx=0 or sinx=1/2. |dw:1374732465835:dw|

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