Tìm tập xác định của hàm số:
a) $y=\sqrt{\sqrt{2} -\frac{1}{\cos 2x}  } $                  b) $y=\frac{3}{\sqrt{\sin 2x -\sin 4x} } $ trên $(0;\pi)$
a) Hàm số xác định khi   $\sqrt{2} \geq  \frac{1}{\cos 2x}                                    (1)  $
*  Nếu $\cos 2x >0$  thì  $(1)          \Leftrightarrow  \cos 2x  \geq  \frac{\sqrt{2} }{2} $
Suy ra $-\frac{\pi}{4}+k2\pi \leq  \frac{\pi}{4}+k2\pi \Leftrightarrow -\frac{\pi}{8}+k\pi \leq  x \leq  \frac{\pi}{4} + k\pi      $
*  Nếu $\cos 2x < 0 $ thì $(1)$ luôn luôn nghiệm đúng.
Vậy tập xác định của hàm số là   $[-\frac{\pi}{8}+k\pi; \frac{\pi}{8}+k \pi]  $

b) Hàm số xác định khi  $\sin 2x -\sin 4x > 0$
                                  $\Leftrightarrow  \sin 4x -\sin 2x <0  \Leftrightarrow   2\sin x \cos 3x <0$
Trên khoảng $(0; \pi)$ thì $\sin x>0$    nên $(1) \Leftrightarrow   \cos 3x <0$
                                  $\Leftrightarrow   \frac{\pi}{2} < 3x < \pi   \Leftrightarrow   \frac{\pi}{6}< x< \frac{\pi}{3}  $.
Vậy tập xác định của hàm số là $(\frac{\pi}{6}+k\pi; \frac{\pi}{3}+k\pi  )$

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