What is the general solution for (sin(2x)-1)(2sin(2x)-1)=0 ?

The general solution for (sin(2x)-1)(2sin(2x)-1)=0 is x= pi/4+pin,x= pi/(12)+pin,x=(5pi)/(12)+pin

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(sin(2x)-1)(2sin(2x)-1)=0

(sin (2 x) - 1) (2 sin (2 x) - 1) = 0

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

Solution steps

(sin (2 x) - 1) (2 sin (2 x) - 1) = 0

Solving each part separately

sin (2 x) - 1 = 0 or 2 sin (2 x) - 1 = 0

sin (2 x) - 1 = 0 : x = \frac{π}{4} + πn

2 sin (2 x) - 1 = 0 : x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Combine all the solutions

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

2 cos^{2} (x) - cos (x) = 2 - sec (x)

sin^{4} (x) - cos^{4} (x) = \frac{1}{2}

x, y = tan (x)

sin (θ) = - 2

\frac{cos ( \frac{x}{2} )}{2} = 0

What is the general solution for (sin(2x)-1)(2sin(2x)-1)=0 ?
The general solution for (sin(2x)-1)(2sin(2x)-1)=0 is x= pi/4+pin,x= pi/(12)+pin,x=(5pi)/(12)+pin