How do I do this?

**iluvmathbutitshard** · Nov 24th 2009, 08:02 AM

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?

**skeeter** · Nov 24th 2009, 08:22 AM

Originally Posted by iluvmathbutitshard

I have to find all solutions of the equation in the interval [0, 2pi) for:
sin 2x = (- sqrt3 / 2)

I am not sure to do this.
Any help is appreciated.

$0 \le x < 2\pi$

$0 \le 2x < 4\pi$

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

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

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

solve for $x$

**iluvmathbutitshard** · Nov 24th 2009, 08:27 AM

so:
x = 5pi/6, 4pi/6, 10pi/6, 11pi/6

Is this right?

**skeeter** · Nov 24th 2009, 08:33 AM

Originally Posted by iluvmathbutitshard

so:
x = 5pi/6, 4pi/6, 10pi/6, 11pi/6

Is this right?

reduce the fractions that can be reduced.

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