Find the general solution of the equation

2[sin(x)]^3=sinx

thx :)

... 2sin^3(x) = sin(x)
or sin(x) ( 2 sin^2(x) - 1 ) = 0
or sin(x) = 0 .... → x = nπ
or sin(x) = 1/√2 → x = { π/4 (8n+1), π/4 (8n+3) }

2[sin(x)]^3/sinx=1

2(sinx)^2=1

(sinx)^2=1/2

sinx=sq rt (1/2)