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Frequently Asked Questions (FAQ)
What is the general solution for sin(θ)sin(2θ)=0,0<= θ<= 2pi ?
- The general solution for sin(θ)sin(2θ)=0,0<= θ<= 2pi is θ=0,θ=pi,θ=2pi,θ= pi/2 ,θ=(3pi)/2