GIẢI PT : a) cosx.cos3x – cos4x = 0 b) sin ²x + sin ²2x + sin ²3x = 3/2 06/12/2021 Bởi Adalyn GIẢI PT : a) cosx.cos3x – cos4x = 0 b) sin ²x + sin ²2x + sin ²3x = 3/2
$\begin{array}{l}a)\quad \cos x\cos3x – \cos4x = 0\\ \Leftrightarrow \dfrac{1}{2}(\cos4x + \cos2x) -\cos4x = 0\\ \Leftrightarrow \dfrac{1}{2}\cos2x – \dfrac{1}{2}\cos4x = 0\\ \Leftrightarrow \cos4x = \cos2x\\ \Leftrightarrow \left[\begin{array}{l}4x = 2x + k2\pi\\4x = -2x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x= k\dfrac{\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\ b) \quad \sin^2x + \sin^22x + \sin^33x = \dfrac{3}{2}\\ \Leftrightarrow \dfrac{1 – \cos2x}{2} + \dfrac{1-\cos4x}{2} + \dfrac{1- \cos6x}{2} = 3\\ \Leftrightarrow \cos2x + \cos6x + \cos4x = 0\\ \Leftrightarrow 2\cos4x\cos2x + \cos4x = 0\\ \Leftrightarrow \cos4x(2\cos2x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos4x = 0\\\cos2x = -\dfrac12\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}4x = \dfrac{\pi}{2} +k\pi\\2x = \pm\dfrac{2\pi}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} +k\dfrac{\pi}{4}\\x = \pm\dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z) \end{array}$ Bình luận
$\begin{array}{l}a)\quad \cos x\cos3x – \cos4x = 0\\ \Leftrightarrow \dfrac{1}{2}(\cos4x + \cos2x) -\cos4x = 0\\ \Leftrightarrow \dfrac{1}{2}\cos2x – \dfrac{1}{2}\cos4x = 0\\ \Leftrightarrow \cos4x = \cos2x\\ \Leftrightarrow \left[\begin{array}{l}4x = 2x + k2\pi\\4x = -2x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x= k\dfrac{\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\ b) \quad \sin^2x + \sin^22x + \sin^33x = \dfrac{3}{2}\\ \Leftrightarrow \dfrac{1 – \cos2x}{2} + \dfrac{1-\cos4x}{2} + \dfrac{1- \cos6x}{2} = 3\\ \Leftrightarrow \cos2x + \cos6x + \cos4x = 0\\ \Leftrightarrow 2\cos4x\cos2x + \cos4x = 0\\ \Leftrightarrow \cos4x(2\cos2x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos4x = 0\\\cos2x = -\dfrac12\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}4x = \dfrac{\pi}{2} +k\pi\\2x = \pm\dfrac{2\pi}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} +k\dfrac{\pi}{4}\\x = \pm\dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z) \end{array}$