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Frequently Asked Questions (FAQ)
What is the general solution for 2sin^2(θ)+cos(θ)=1,0<= θ<= 2pi ?
- The general solution for 2sin^2(θ)+cos(θ)=1,0<= θ<= 2pi is θ=(2pi)/3 ,θ=(4pi)/3 ,θ=0,θ=2pi