Solving Tri equation

**Neconine** · December 7th 2009, 09:43 AM

Having trouble with this question.

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

**Amer** · December 7th 2009, 10:37 AM

Originally Posted by Neconine

Having trouble with this question.

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

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

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

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

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

in general

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

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

or

$3x = 2\pi - \frac{\pi}{3} + 2n\pi$ n is like above

**Lucio Carvalho** · December 7th 2009, 11:39 AM

Hello Neconine,
I agree with Amer.
The attachment shows one of the possible ways to solve the equation.
I hope you understand!
Bye!

**Neconine** · December 7th 2009, 06:26 PM

Thanks to the both of you.

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

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