What is the general solution for 2sin^2(2x)-sin(2x)-3=0 ?

The general solution for 2sin^2(2x)-sin(2x)-3=0 is x=(3pi)/4+pin

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2sin^2(2x)-sin(2x)-3=0

2 sin^{2} (2 x) - sin (2 x) - 3 = 0

x = \frac{3 π}{4} + πn

Solution steps

2 sin^{2} (2 x) - sin (2 x) - 3 = 0

Solve by substitution

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

sin (2 x) = \frac{3}{2} :

No Solution

sin (2 x) = - 1 : x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{3 π}{4} + πn

cos (π + x) + cos (π + x) = 1

1 + 2 sin (θ) = 0

x, z = arcsin (x y)

2 cos (x) - 2 = 0

4 sin^{2} (θ) - 13 sin (θ) + 9 = 0

What is the general solution for 2sin^2(2x)-sin(2x)-3=0 ?
The general solution for 2sin^2(2x)-sin(2x)-3=0 is x=(3pi)/4+pin