Question

Help please!!!!!!!!!!!!

Answer 1

Answer:

$\frac{\pi }{3},\:\frac{2\pi }{3}$

Step-by-step explanation:

Let a=THETA (angle)

$1+\sin \left(a\right)-1=\frac{\sqrt{3}+2}{2}-1\\\sin \left(a\right)=\frac{\sqrt{3}+2}{2}-1\\a=\arcsin \left(\frac{\sqrt{3}+2}{2}-1\right)+2\pi n,\:a=\pi -\arcsin \left(\frac{\sqrt{3}+2}{2}-1\right)+2\pi n\\a=\frac{\pi }{3},\:a=\frac{2\pi }{3}$