Giúp vs.Mk cần gấp lắm y=3-2IsinxI y= cosx + cos(x-π/3) y=2-(√cosx) y= sin2x + 1 y= cos(x-π/6)

0 bình luận về “Giúp vs.Mk cần gấp lắm y=3-2IsinxI y= cosx + cos(x-π/3) y=2-(√cosx) y= sin2x + 1 y= cos(x-π/6)”

1. Đáp án:

   Giải thích các bước giải:
   \[\begin{array}{l}
   a.0 \le |\sin x| \le 1 = > 1 \le y \le 3 = > \min y = 1 \Leftrightarrow x = \frac{\pi }{2} + k\pi ;\max y = 3 \Leftrightarrow x = k\pi \\
   b.y = \cos x + \cos (x – \frac{\pi }{3}) = 2\cos (x – \frac{\pi }{6}).\cos \frac{\pi }{6} = \sqrt 3 \cos (x – \frac{\pi }{6})\\
   – 1 \le \cos (x – \frac{\pi }{6}) \le 1 = > – \sqrt 3 \le y \le \sqrt 3 = > \min y = – \sqrt 3 \Leftrightarrow x = \frac{{7\pi }}{6} + k2\pi ;\max y = \sqrt 3 \Leftrightarrow x = \frac{\pi }{6} + k2\pi \\
   c.y = 2 – \sqrt {\cos x} \\
   dk:\cos x \ge 0\\
   0 \le \cos x \le 1 = > 1 \le y \le 2 = > \min y = 1 \Leftrightarrow x = k2\pi ;\max y = 2 \Leftrightarrow x = \frac{\pi }{2} + k2\pi \\
   d.y = \sin 2x + 1\\
   – 1 \le \sin 2x \le 1 = > 0 \le y \le 2 = > \min y = 0;\max y = 2\\
   e.y = \cos (x – \frac{\pi }{6})\\
   – 1 \le \cos (x – \frac{\pi }{6}) \le 1 = > \min y = – 1;\max y = 1
   \end{array}\]

   Trả lời