What is the general solution for 2sin^2(2θ)+5sin(2θ)=3 ?

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

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2sin^2(2θ)+5sin(2θ)=3

2 sin^{2} (2 θ) + 5 sin (2 θ) = 3

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

Solution steps

2 sin^{2} (2 θ) + 5 sin (2 θ) = 3

Solve by substitution

sin (2 θ) = \frac{1}{2}, sin (2 θ) = - 3

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

sin (2 θ) = - 3 :

No Solution

Combine all the solutions

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

3 cot (x) + 4 = 7

2 sin (4 θ) + 1 = 0

cot^{2} (x) csc^{2} (x) + 2 csc^{2} (x) - cot^{2} (x) = 2

cos (t) = \frac{3}{5}, 0 < t < \frac{π}{2}

cos (x) = \frac{- 2}{2}

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