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

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

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

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

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

Solution steps

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

Rewrite using trig identities

- 2 sin (2 x) + 1 = 0

Move

1

to the right side

- 2 sin (2 x) = - 1

Divide both sides by

- 2

sin (2 x) = \frac{1}{2}

General solutions for

sin (2 x) = \frac{1}{2}

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

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

2 sin^{3} (x) + sin^{2} (x) - 2 sin (x) - 1 = 0

5 cos^{2} (θ) + cos (θ) = 0

x, z = sin (x y)

3 sec (x) + 3 tan (x) = 3

tan (x) + cot (x) = 4 sin (2 x)

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