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

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

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1-sin(2x)=0

1 - sin (2 x) = 0

x = \frac{π}{4} + πn

Solution steps

1 - sin (2 x) = 0

Move

1

to the right side

- sin (2 x) = - 1

Divide both sides by

- 1

sin (2 x) = 1

General solutions for

sin (2 x) = 1

2 x = \frac{π}{2} + 2 πn

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

cos (p) - sin (p) = 1

x, y = 2 sin (x)

2 cos (3 x) + cos (2 x) + 1 = 0

4 + 2 tan^{2} (θ) = 5 tan^{2} (θ) + 3

sin (x) = sin (\frac{π}{5})

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