Tentukan nilai x yang memenuhi persamaan:
\(\sin 2x = 0,5\) dimana \(0 \leq x \leq 2π\)
\begin{equation*}
\begin{split}
& \sin 2x = 0,5 \\\\
& \sin 2x = \sin \frac{1}{6} \pi
\end{split}
\end{equation*}
Solusi 1
\begin{equation*}
\begin{split}
2x & = \frac{1}{6} \pi + k \:.\: 2 \pi \\\\
x & = \frac{1}{12} \pi + k \:.\: \pi \\\\
k & = 0 \rightarrow {\color {red} x = \frac{1}{12} \pi} \\\\
k & = 1 \rightarrow {\color {red} x = 1\frac{1}{12} \pi}
\end{split}
\end{equation*}
Solusi 2
\begin{equation*}
\begin{split}
2x & = \left(\pi - \frac{1}{6} \pi \right) + k \:.\: 2 \pi \\\\
2x & = \frac{5}{6} \pi + k \:.\: 2 \pi \\\\
x & = \frac{5}{12} \pi + k \:.\: \pi \\\\
k & = 0 \rightarrow {\color {red} x = \frac{5}{12} \pi} \\\\
k & = 1 \rightarrow {\color {red} x = 1\frac{5}{12} \pi}
\end{split}
\end{equation*}
HP = \(\left\{\frac{1}{12} \pi, \frac{5}{12} \pi, 1\frac{1}{12} \pi, 1\frac{5}{12} \pi\right\}\)