Let's address each part of the problem step by step:

A) Given: \(\sin x = \sin\left(\frac{2\pi}{3}\right)\).

The sine function is periodic with a period of \(2\pi\), and it has the property that \(\sin\theta = \sin(\pi - \theta)\). Therefore, the solutions for \(x\) are:

1. \(x = \frac{2\pi}{3} + 2k\pi\), where \(k\) is any integer.
2. \(x = \pi - \frac{2\pi}{3} + 2k\pi = \frac{\pi}{3} + 2k\pi\), where \(k\) is any integer.

B) Given: \(\sin(2x + 20^\circ) = \sin(-60^\circ)\).

Using the property that \(\sin A = \sin B\) implies \(A = B + 2k\pi\) or \(A = \pi - B + 2k\pi\), we have:

1. \(2x + 20^\circ = -60^\circ + 360^\circ k\)
\[
2x = -80^\circ + 360^\circ k
\]
\[
x = -40^\circ + 180^\circ k
\]

2. \(2x + 20^\circ = 180^\circ + 60^\circ + 360^\circ k = 240^\circ + 360^\circ k\)
\[
2x = 220^\circ + 360^\circ k
\]
\[
x = 110^\circ + 180^\circ k
\]

where \(k\) is any integer.

C) This part is confusing because the notation is not clear. However, if you rewrite it to involve \(\sin 2x = -\frac{1}{2}\), then using the sine property:

\[
2x = \pi + \frac{\pi}{6} + 2k\pi \text{ or } 2x = 2\pi - \frac{\pi}{6} + 2k\pi
\]

1. \(2x = \frac{7\pi}{6} + 2k\pi \Rightarrow x = \frac{7\pi}{12} + k\pi\)
2. \(2x = \frac{11\pi}{6} + 2k\pi \Rightarrow x = \frac{11\pi}{12} + k\pi\)

D) Given: \(\sin(x + 45^\circ) = -\frac{\sqrt{2}}{2}\).

The sine function equals \(-\frac{\sqrt{2}}{2}\) at angles \(225^\circ\) (which is \(180^\circ + 45^\circ\)) and \(315^\circ\) (which is \(360^\circ - 45^\circ\)). So:

1. \(x + 45^\circ = 225^\circ + 360^\circ k\)
\[
x = 180^\circ k
\]

2. \(x + 45^\circ = 315^\circ + 360^\circ k\)
\[
x = 270^\circ + 360^\circ k
\]

where \(k\) is any integer.

Overall, each statement provides general solutions for \(x\) based on the periodic properties of the sine function, considering all possible angles within a full cycle of the unit circle.