Đáp án:
\(\begin{cases}\min y = 1\Leftrightarrow x = - \dfrac{\pi}{6} + \dfrac{k\pi}{2}\\
\max y = 5\Leftrightarrow x = \dfrac{\pi}{12} + \dfrac{k\pi}{2}\\
\end{cases}\quad (k\in\Bbb Z)\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad y = 4(\sin^4x + \cos^4x) +\sqrt3\sin4x\\
\Leftrightarrow y =4[(\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x] + \sqrt3\sin4x\\
\Leftrightarrow y = 4\left(1 - \dfrac12\sin^22x\right) + \sqrt3\sin4x\\
\Leftrightarrow y = -2\sin^22x + \sqrt3\sin4x + 4\\
\Leftrightarrow y = -2\cdot \dfrac{1 - \cos4x}{2}+ \sqrt3\sin4x + 4\\
\Leftrightarrow y = \sqrt3\sin4x + \cos4x+3\\
\Leftrightarrow y = 2\left(\dfrac{\sqrt3}{2}\sin4x + \dfrac12\cos4x\right) + 3\\
\Leftrightarrow y = 2\sin\left(4x + \dfrac{\pi}{6}\right) + 3\\
\text{Ta có:}\\
\quad - 1 \leqslant \sin\left(4x + \dfrac{\pi}{6}\right) \leqslant 1\\
\Leftrightarrow -2 \leqslant 2\sin\left(4x + \dfrac{\pi}{6}\right)\leqslant 2\\
\Leftrightarrow 1 \leqslant 2\sin\left(4x + \dfrac{\pi}{6}\right) + 3 \leqslant 5\\
\text{Hay}\ \ 1 \leqslant y \leqslant 5\\
\text{Do đó:}\\
\bullet\quad \min y = 1\\
\Leftrightarrow \sin\left(4x + \dfrac{\pi}{6}\right) = -1\\
\Leftrightarrow 4x + \dfrac{\pi}{6} = - \dfrac{\pi}{2} + k2\pi\\
\Leftrightarrow x = - \dfrac{\pi}{6} + \dfrac{k\pi}{2}\quad (k\in\Bbb Z)\\
\bullet\quad \max y = 5\\
\Leftrightarrow \sin\left(4x + \dfrac{\pi}{6}\right) = 1\\
\Leftrightarrow 4x + \dfrac{\pi}{6} = \dfrac{\pi}{2} +k2\pi\\
\Leftrightarrow x = \dfrac{\pi}{12} + \dfrac{k\pi}{2}\quad (k\in\Bbb Z)\\
\end{array}\)
