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