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

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## 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?

- Nov 24th 2009, 08:22 AM #2
$\displaystyle 0 \le x < 2\pi$

$\displaystyle 0 \le 2x < 4\pi$

values in the interval $\displaystyle [0, 4\pi)$ where

$\displaystyle \sin(2x) = -\frac{\sqrt{3}}{2}$ ...

$\displaystyle 2x = \frac{4\pi}{3} \, , \, \frac{5\pi}{3} \, , \, \frac{10\pi}{3} \, , \, \frac{11\pi}{3}$

solve for $\displaystyle x$

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

- Joined
- Nov 2009
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- Nov 24th 2009, 08:33 AM #4