giải pt 4 vs https://img.hoidap247.com/picture/question/20201006/tini_1601991777923.jpg

$\begin{array}{l}\cos^23x – \sin^23x = \cos x\\ \Leftrightarrow \cos6x = \cos x \qquad \text{(Áp dụng công thức $\cos^2a – \sin^2a = \cos2a$ với a = 3x)}\\ \Leftrightarrow \left[\begin{array}{l}6x = x + k2\pi\\6x = – x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\dfrac{2\pi}{5}\\x = k\dfrac{2\pi}{7}\end{array}\right.\quad (k \in \Bbb Z)\\ \text{Hoặc áp dụng công thức hạ bậc, dài hơn}\\ \cos^23x – \sin^23x = \cos x\\ \Leftrightarrow \dfrac{1 + \cos6x}{2} – \dfrac{1 – \cos6x}{2} = \cos x\\ \Leftrightarrow 1 + \cos6x – (1 – \cos6x) = 2\cos x\\ \Leftrightarrow 2\cos6x = 2\cos x\\ \Leftrightarrow \cos6x = \cos x\\ \Leftrightarrow \left[\begin{array}{l}x = k\dfrac{2\pi}{5}\\x = k\dfrac{2\pi}{7}\end{array}\right.\quad (k \in \Bbb Z)\\ \end{array}$