[Toán 11] Lượng giác

N

niemkieuloveahbu

dễ thấy [TEX]sin x > 0 [/TEX]
[TEX]16 sin^4 x + 5 = 6 \sqrt[6]{ sin^2 x ( 4 sin^2 x+1)^2} = 3 . \sqrt[6]{8 sin^2 x ( 4sin^2 x +1)^2 . 2. 2 .2 }\\ \le \frac12 ( 8 sin^2 x + 4 sin^2 x + 1 + 4 sin^2 x +1 +2 +2 + 2) = 8 sin^2 x + 4\\ \Leftrightarrow ( 4 sin^2 x -1)^2 \le 0\\ \Leftrightarrow sin^2 x = \frac12[/TEX]
 
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