fiind all solutions at interval [0,2pi] sin2x = 0

Answer 1

We want to find the solutions to

sin(2x) = 0

where x varies over the interval from 0 to 2π.

Sin(y) is 0.0 at 0, π, 2π, 3π, 4π, etc.

The argument of sin(2x) varies from 0 to 4π, so we have the following solutions:

2x = 0 => x = 0

2x = π => x = π/2

2x = 2π => x = π

2x = 3π => x = (3/2)π

and

2x = 4π => x = 2π

Answer 2

The angles 0, pi and 2pi have a sin of 0

so 2x=0 x=0

2x=pi x=pi/2

2x=2pi x=pi

So solutions are 0,pi/2 and pi