Tentukan nilai x yang memenuhi persamaan:
\(2\cos^2 x + \cos x - 1 = 0, -360^{\text{o}} \leqslant x \leqslant 360^{\text{o}} \)
\begin{equation*}
\begin{split}
& 2\cos^2 x + \cos x - 1 = 0 \\\\
& (2\cos x -1)(\cos x + 1) = 0 \\\\
& 2\cos x - 1 = 0 \: \text{atau} \: \cos x + 1 = 0 \\\\
& \cos x = \tfrac{1}{2} \: \text{atau} \: \cos x = -1
\end{split}
\end{equation*}
Untuk \(\cos x = \frac{1}{2}\)
\begin{equation*}
\begin{split}
& \cos x = \tfrac{1}{2} \\\\
& \cos x = \cos 60^{\text{o}}
\end{split}
\end{equation*}
Solusi 1
\begin{equation*} \begin{split} x & = 60^{\text{o}} + k \:.\: 360^{\text{o}} \\\\ k & = -1 \rightarrow {\color {red} x = - 300^{\text{o}}} \\\\ k & = 0 \rightarrow {\color {red} x = 60^{\text{o}}} \end{split} \end{equation*}
Solusi 2
\begin{equation*} \begin{split} x & = - 60^{\text{o}} + k \:.\: 360^{\text{o}} \\\\ k & = 0 \rightarrow {\color {red} x = - 60^{\text{o}}} \\\\ k & = 1 \rightarrow {\color {red} x = 300^{\text{o}}} \end{split} \end{equation*}
Untuk \(\cos x = -1\)
\begin{equation*}
\begin{split}
& \cos x = -1 \\\\
& \cos x = \cos 180^{\text{o}}
\end{split}
\end{equation*}
Solusi 1
\begin{equation*} \begin{split} x & = 180^{\text{o}} + k \:.\: 360^{\text{o}} \\\\ k & = -1 \rightarrow {\color {red} x = - 180^{\text{o}}} \\\\ k & = 0 \rightarrow {\color {red} x = 180^{\text{o}}} \end{split} \end{equation*}
Solusi 2
\begin{equation*} \begin{split} x & = - 180^{\text{o}} + k \:.\: 360^{\text{o}} \\\\ k & = 0 \rightarrow {\color {red} x = - 180^{\text{o}}} \\\\ k & = 1 \rightarrow {\color {red} x = 180^{\text{o}}} \end{split} \end{equation*}
HP = \(\{ -300^{\text{o}} , -180^{\text{o}}, -60^{\text{o}}, 60^{\text{o}}, 180^{\text{o}}, 300^{\text{o}} \}\)