Having trouble with this question.
Determine exact solutions for sin3x + square root 3 / 2 =0, where 0 <= theta <= 2 pi
$\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