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

The general solution for 2sin^2(2θ)=-1+3sin(2θ) is θ= pi/4+pin,θ= pi/(12)+pin,θ=(5pi)/(12)+pin

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2sin^2(2θ)=-1+3sin(2θ)

2 sin^{2} (2 θ) = - 1 + 3 sin (2 θ)

θ = \frac{π}{4} + πn, θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

Solution steps

2 sin^{2} (2 θ) = - 1 + 3 sin (2 θ)

Solve by substitution

sin (2 θ) = 1, sin (2 θ) = \frac{1}{2}

sin (2 θ) = 1 : θ = \frac{π}{4} + πn

sin (2 θ) = \frac{1}{2} : θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

6 tan^{2} (θ) + 10 tan (θ) = 9 tan (θ) + 2

y, z = cos (y)

cos (x) = \frac{64}{83}

2 sin (z) = 1

cos (\frac{π}{2} - θ) = - sin (θ)

What is the general solution for 2sin^2(2θ)=-1+3sin(2θ) ?
The general solution for 2sin^2(2θ)=-1+3sin(2θ) is θ= pi/4+pin,θ= pi/(12)+pin,θ=(5pi)/(12)+pin