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

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

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

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

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

Solution steps

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

Solving each part separately

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

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

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

Combine all the solutions

x = πn, x = \frac{π}{2} + πn, x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Since the equation is undefined for:

πn, \frac{π}{2} + πn

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

tan (x - π) + 2 sin (x - π) = 0

tan (4 5^{\circ}) = \frac{1 - cos ( u )}{sin ( u )}

1 - 2 cos^{2} (x) = 0

4 sin (x) = 1

sin (x + 1) = 0

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