Giải phương trình lượng giác cơ bản

a) \(\cos\left(3x - \dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}\) Nhận xét: \(-\dfrac{\sqrt{2}}{2} = \cos\dfrac{3\pi}{4}\), nên: \(\cos\left(3x - \dfrac{\pi}{4}\right) = \cos\dfrac{3\pi}{4}\) \(\Leftrightarrow \left[\begin{array}{l} 3x - \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi \\[6pt] 3x - \dfrac{\pi}{4} = -\dfrac{3\pi}{4} + k2\pi \end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l} 3x = \pi + k2\pi \\[6pt] 3x = -\dfrac{\pi}{2} + k2\pi \end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l} x = \dfrac{\pi}{3} + \dfrac{k2\pi}{3} \\[6pt] x = -\dfrac{\pi}{6} + \dfrac{k2\pi}{3} \end{array}\right. \quad (k \in \mathbb{Z})\) b) \(2\sin^2 x - 1 + \cos 3x = 0\) Dùng đồng nhất thức \(2\sin^2 x - 1 = -\cos 2x\): \(-\cos 2x + \cos 3x = 0\) \(\Leftrightarrow \cos 3x = \cos 2x\) \(\Leftrightarrow \left[\begin{array}{l} 3x = 2x + k2\pi \\[6pt] 3x = -2x + k2\pi \end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l} x = k2\pi \\[6pt] 5x = k2\pi \end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l} x = k2\pi \\[6pt] x = \dfrac{k2\pi}{5} \end{array}\right. \quad (k \in \mathbb{Z})\) c) \(\tan\left(2x + \dfrac{\pi}{5}\right) = \tan\left(x - \dfrac{\pi}{6}\right)\) Áp dụng công thức nghiệm tan: \(2x + \dfrac{\pi}{5} = x - \dfrac{\pi}{6} + k\pi\) \(\Leftrightarrow x = -\dfrac{\pi}{6} - \dfrac{\pi}{5} + k\pi\) \(\Leftrightarrow x = -\dfrac{5\pi}{30} - \dfrac{6\pi}{30} + k\pi\) \(\Leftrightarrow x = -\dfrac{11\pi}{30} + k\pi \quad (k \in \mathbb{Z})\)