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Determine exact solutions for sin3x + square root 3 / 2 =0, where 0 <= theta <= 2 pi

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- Dec 7th 2009, 09:43 AMNeconineSolving Tri equation
Having trouble with this question.

Determine exact solutions for sin3x + square root 3 / 2 =0, where 0 <= theta <= 2 pi - Dec 7th 2009, 10:37 AMAmer

$\displaystyle \sin 3x + \frac{\sqrt{3}}{2} = 0 $

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

$\displaystyle 3x = \pi + \frac{\pi}{3} \Rightarrow x = \frac{4\pi}{9} $

$\displaystyle 3x = 2\pi - \frac{\pi}{3} \Rightarrow x = \frac{5\pi}{9} $

in general

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

$\displaystyle 3x = \pi + \frac{\pi}{3} + 2n\pi $ n natural number 0,1,2,3,... take the numbers such that $\displaystyle 0\leq x \leq 2 \pi $

or

$\displaystyle 3x = 2\pi - \frac{\pi}{3} + 2n\pi $ n is like above - Dec 7th 2009, 11:39 AMLucio Carvalhohelp for the exercise
Hello Neconine,

I agree with Amer.

The attachment shows one of the possible ways to solve the equation.

I hope you understand!

Bye! - Dec 7th 2009, 06:26 PMNeconine
Thanks to the both of you.

You have been a great help, I really appreciate this.