Đáp án:
\(S = \left\{\dfrac{\pi}{4} + k\dfrac{\pi}{2};\ \dfrac{\pi}{6} + k\pi;\ -\dfrac{\pi}{6} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \cos^6x - \sin^6x = \dfrac{13}{8}\cos^22x\\
\Leftrightarrow (\cos^2x - \sin^2x)(\cos^4x + \cos^2x.\sin^2x + \sin^4x) = \dfrac{13}{8}\cos^22x\\
\Leftrightarrow \cos2x\left[(\cos^2x + \sin^2x)^2 - \cos^2x.\sin^2x\right] -\dfrac{13}{8}\cos^22x = 0\\
\Leftrightarrow \cos2x\left(1 - \dfrac14\sin^22x\right) - \dfrac{13}{8}\cos^22x = 0\\
\Leftrightarrow \cos2x(8 - 2\sin^22x - 13\cos2x) = 0\\
\Leftrightarrow \cos2x[8 - 2(1-\cos^22x) - 13\cos2x] = 0\\
\Leftrightarrow \cos2x(2\cos^22x - 13\cos2x +6) = 0\\
\Leftrightarrow \cos2x(2\cos2x - 1)(\cos2x - 6) = 0\\
\Leftrightarrow \left[\begin{array}{l}\cos2x = 0\\\cos2x = \dfrac12\\cos2x = 6\quad (vn)\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{2} + k\pi\\2x = \dfrac{\pi}{3} + k2\pi\\2x = -\dfrac{\pi}{3} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x = \dfrac{\pi}{6} + k\pi\\x = -\dfrac{\pi}{6} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{4} + k\dfrac{\pi}{2};\ \dfrac{\pi}{6} + k\pi;\ -\dfrac{\pi}{6} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
