Đáp án:
2sin(($\frac{\pi}{5}$ - x) = 1
⇔ sin($\frac{\pi}{5}$ - x) = $\frac{1}{2}$
⇔ sin($\frac{\pi}{5}$ - x) = sin$\frac{\pi}{6}$
⇔ \(\left[ \begin{array}{l}\frac{\pi}{5} - x = \frac{\pi}{6} + k2\pi \\\frac{\pi}{5} - x = \pi-\frac{\pi}{6} + k2\pi \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = \frac{\pi}{30} - k2\pi \\x =\frac{-19\pi}{30} - k2\pi \end{array} \right.\)