Giải phương trình lượng giác dạng tổng sin và cos

a) Dùng công thức \(\cos 4x = 1 - 2\sin^2 2x\): \(\sin 2x + \cos 4x = 0 \Leftrightarrow \sin 2x + 1 - 2\sin^2 2x = 0\) \(\Leftrightarrow 2\sin^2 2x - \sin 2x - 1 = 0\) \(\Leftrightarrow \left[\begin{array}{l}\sin 2x = 1\\\sin 2x = -\dfrac{1}{2}\end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l}\sin 2x = \sin\dfrac{\pi}{2}\\\sin 2x = \sin\left(-\dfrac{\pi}{6}\right)\end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{2} + k2\pi\\2x = -\dfrac{\pi}{6} + k2\pi\\2x = \pi + \dfrac{\pi}{6} + k2\pi\end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{12} + k\pi\\x = \dfrac{7\pi}{12} + k\pi\end{array}\right. \quad (k \in \mathbb{Z})\) b) Chuyển vế rồi dùng công thức tổng thành tích: \(\cos 3x = -\cos 7x \Leftrightarrow \cos 3x + \cos 7x = 0\) \(\Leftrightarrow 2\cos 5x \cos 2x = 0\) \(\Leftrightarrow \left[\begin{array}{l}\cos 5x = 0\\\cos 2x = 0\end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l}5x = \dfrac{\pi}{2} + k\pi\\2x = \dfrac{\pi}{2} + k\pi\end{array}\right.\) \(\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{10} + k\dfrac{\pi}{5}\\x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\end{array}\right. \quad (k \in \mathbb{Z})\)