Tìm max, min

N

nguyenbahiep1

Tìm gtln, gtnn của hs sau:
y=[TEX]\sqrt{3} sin2x - cos2x[/TEX]

[TEX]y = 2 ( \frac{\sqrt{3}}{2}.sin 2x - \frac{1}{2}.cos2x) = 2 ( sin 2x . cos ( \frac{\pi}{6}) - sin (\frac{\pi}{6}).cos2x) \\ y = 2.sin ( 2x - \frac{\pi}{6}) \\ -1 \leq sin ( 2x - \frac{\pi}{6}) \leq 1 \Rightarrow max y = 2 \\ min y = - 2[/TEX]
 
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