a. Ta có:
\(\sin 4x = \sin {\pi \over 5} \Leftrightarrow \left[ {\matrix{{4x = {\pi \over 5} + k2\pi } \cr {4x = \pi - {\pi \over 5} + k2\pi } \cr} \,\,\left( {k \in\mathbb Z} \right) \Leftrightarrow \left[ {\matrix{{x = {\pi \over {20}} + k{\pi \over 2}} \cr {x = {\pi \over 5} + k{\pi \over 2}} \cr} } \right.} \right.\,\,\left( {k \in\mathbb Z} \right)\)
b. Vì \( - {1 \over 2} =- \sin {\pi \over 6} = \sin \left( { - {\pi \over 6}} \right)\) nên :\(\sin \left( {{{x + \pi } \over 5}} \right) = - {1 \over 2} \Leftrightarrow \left[ {\matrix{{{{x + \pi } \over 5} = - {\pi \over 6} + k2\pi } \cr {{{x + \pi } \over 5} = \pi + {\pi \over 6} + k2\pi } \cr} } \right. \Leftrightarrow \left[ {\matrix{{x = - {{11\pi } \over 6} + k10\pi } \cr {x = {{29\pi } \over 6} + k10\pi } \cr} } \right.\,\,\left( {k \in\mathbb Z} \right)\)
c.
\(\cos {x \over 2} = \cos \sqrt 2 \Leftrightarrow {x \over 2} = \pm \sqrt 2 + k2\pi \Leftrightarrow x = \pm 2\sqrt 2 + k4\pi \,\left( {k \in\mathbb Z} \right)\)
d. Vì \(0 < {2 \over 5} < 1\) nên có số \(α\) sao cho \(\cos \alpha = {2 \over 5}.\) Do đó :
\(\cos \left( {x + {\pi \over {18}}} \right) = {2 \over 5} \Leftrightarrow \cos \left( {x + {\pi \over {18}}} \right) = \cos \alpha \Leftrightarrow x = \pm \alpha - {\pi \over {18}} + k2\pi ,k \in \mathbb Z\)
