What is the general solution for 2sin^2(θ)+3sin(θ)=4sin(θ)+1 ?

The general solution for 2sin^2(θ)+3sin(θ)=4sin(θ)+1 is θ= pi/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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2sin^2(θ)+3sin(θ)=4sin(θ)+1

2 sin^{2} (θ) + 3 sin (θ) = 4 sin (θ) + 1

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

2 sin^{2} (θ) + 3 sin (θ) = 4 sin (θ) + 1

Solve by substitution

sin (θ) = 1, sin (θ) = - \frac{1}{2}

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

cos (θ - \frac{π}{6}) = \frac{3}{2}

csc (θ) = - \frac{5}{3}

2 sin^{2} (t) cos (t) = 0

2 cos (θ) + sin (θ) = 0

2 sin^{2} (x) + sin (x) - 1 = 0, 0 < x < 2 π

What is the general solution for 2sin^2(θ)+3sin(θ)=4sin(θ)+1 ?
The general solution for 2sin^2(θ)+3sin(θ)=4sin(θ)+1 is θ= pi/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin