Question

Help fast pls! tysm, much appreciated

Answer 1

Answer: b. and c.

Step-by-step explanation:

$\displaystyle\\Let\ arcsin\frac{-\sqrt{3} }{2} =x\ \ \ \ -\frac{\pi }{2} < x < \frac{\pi }{2}\\So,\\sin(arcsin\frac{-\sqrt{3} }{2} )=sin(x)\\\frac{-\sqrt{3} }{2}=sin(x )\\sin(x)=-\frac{\sqrt{3}}{2} \\\\x_1=\frac{4\pi }{3}+2\pi n\\\\ x_2=\frac{5\pi }{3} +2\pi n.\\\\ -\frac{\pi }{2} < x < \frac{\pi }{2},\\\\Hence,\\\\x_1=\frac{4\pi }{3} \\\\x_2=\frac{5\pi }{3} .$