I have to find all solutions of the equation in the interval [0, 2pi) for:

sin 2x = (- sqrt3 / 2)

My work:

sin 2x = -sqrt(3)/2

2x = 5pi/3, 4pi/3

x = 5pi/6 or 2pi/3

Is this correct?

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- November 24th 2009, 08:02 AM #1

- Joined
- Nov 2009
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- 87

## How do I do this?

I have to find all solutions of the equation in the interval [0, 2pi) for:

sin 2x = (- sqrt3 / 2)

My work:

sin 2x = -sqrt(3)/2

2x = 5pi/3, 4pi/3

x = 5pi/6 or 2pi/3

Is this correct?

- November 24th 2009, 08:22 AM #2

- November 24th 2009, 08:27 AM #3

- Joined
- Nov 2009
- Posts
- 87

- November 24th 2009, 08:33 AM #4