Tentukan nilai x yang memenuhi persamaan:
\(\sqrt{3} \cos x - \sin x = 2\) dimana \(0 \leq x \leq 2 \pi\)
Bentuk \(\sqrt{3} \cos x - \sin x = 2 \) diubah menjadi bentuk \(k \cos (x - \alpha) = c\)
\(k = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2 \)
\(\tan \alpha = \dfrac ba = \dfrac{-1}{\sqrt{3}} = -\dfrac{1}{3} \sqrt{3} \)
\( \alpha = \dfrac{1}{6} \pi\)
Karena a (+) dan b (−) maka \(\alpha\) berada di kuadran 4
\(\alpha = 2\pi - \dfrac{1}{6} \pi = \dfrac{11}{12} \pi \)
\begin{equation*}
\begin{split}
\sqrt{3} \cos x - \sin x & = 2 \\\\
2 \cos \left(x - \frac{11}{12} \pi \right) & = 2 \\\\
\cos \left(x - \frac{11}{12} \pi \right) & = 1 \\\\
\cos \left(x - \frac{11}{12} \pi \right) & = \cos 0 \\\\
x - \frac{11}{12} \pi & = 0 + k \:.\: 2\pi \\\\
x & = \frac{11}{12} \pi + k \:.\: 2\pi \\\\
k & = 0 \rightarrow {\color {red} x = \frac{11}{12} \pi }
\end{split}
\end{equation*}
HP = \(\{\frac{11}{12} \pi\}\)