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Solution: The equation $ \sin(2\theta) = \frac{\sqrt{3}}{2} $ implies $ 2\theta = 60^\circ + 360^\circ k $ or $ 2\theta = 120^\circ + 360^\circ k $ for integer $ k $. Solving for $ \theta $, we get $ \theta = 30^\circ + 180^\circ k $ or $ \theta = 60^\circ + 180^\circ k $. Within $ [0^\circ, 360^\circ] $, the solutions are $ 30^\circ, 60^\circ, 210^\circ, 240^\circ $. Thus, the answer is $ \boxed{30^\circ}, \boxed{60^\circ}, \boxed{210^\circ}, \boxed{240^\circ} $.
Question: Determine the matrix $ \mathbf{M} $ such that $ \mathbf{M} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} $.
Solution: Let $ \mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $. Multiplying $ \mathbf{M} $ with $ \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} $ gives:
\begin{pmatrix} a(1) + b(3) & a(2) + b(4) \\ c(1) + d(3) & c(2) + d(4) \end{pmatrix} = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}.
This leads to the system:
$ a + 3b = 5 $,
$ 2a + 4b = 6 $,
$ c + 3d = 7 $,
$ 2c + 4d = 8 $.
Solving equations 1 and 2: From 1, $ a = 5 - 3b $. Substituting into 2: $ 2(5 - 3b) + 4b = 6 \Rightarrow 10 - 6b + 4b = 6 \Rightarrow -2b = -4 \Rightarrow b = 2 $. Then $ a = 5 - 6 = -1 $.