Giải phương trình lượng giác cơ bản

a) \(\sin x = \dfrac{\sqrt{3}}{2}\) \(\Leftrightarrow \sin x = \sin\dfrac{\pi}{3}\) \(\Leftrightarrow \left[\begin{array}{l} x = \dfrac{\pi}{3} + k2\pi \\ x = \pi - \dfrac{\pi}{3} + k2\pi \end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l} x = \dfrac{\pi}{3} + k2\pi \\ x = \dfrac{2\pi}{3} + k2\pi \end{array}\right. \quad (k \in \mathbb{Z})\) b) \(2\cos x = -\sqrt{2}\) \(\Leftrightarrow \cos x = -\dfrac{\sqrt{2}}{2}\) \(\Leftrightarrow \cos x = \cos\dfrac{3\pi}{4}\) \(\Leftrightarrow \left[\begin{array}{l} x = \dfrac{3\pi}{4} + k2\pi \\ x = -\dfrac{3\pi}{4} + k2\pi \end{array}\right. \quad (k \in \mathbb{Z})\) c) \(\sqrt{3}\tan\left(\dfrac{x}{2} + 15^o\right) = 1\) \(\Leftrightarrow \tan\left(\dfrac{x}{2} + \dfrac{\pi}{12}\right) = \dfrac{1}{\sqrt{3}}\) \(\Leftrightarrow \tan\left(\dfrac{x}{2} + \dfrac{\pi}{12}\right) = \tan\dfrac{\pi}{6}\) \(\Leftrightarrow \dfrac{x}{2} + \dfrac{\pi}{12} = \dfrac{\pi}{6} + k\pi\) \(\Leftrightarrow \dfrac{x}{2} = \dfrac{\pi}{12} + k\pi\) \(\Leftrightarrow x = \dfrac{\pi}{6} + 2k\pi \quad (k \in \mathbb{Z})\) d) \(\cot(2x - 1) = \cot\dfrac{\pi}{5}\) \(\Leftrightarrow 2x - 1 = \dfrac{\pi}{5} + k\pi\) \(\Leftrightarrow 2x = 1 + \dfrac{\pi}{5} + k\pi\) \(\Leftrightarrow x = \dfrac{1}{2} + \dfrac{\pi}{10} + \dfrac{k\pi}{2} \quad (k \in \mathbb{Z})\)