[Toán 11] Tìm GTLN, GTNN

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nguyenbahiep1

[laTEX]y = \frac{sin 2x + cos2x+1}{2-cos2x} \\ \\ 2y - y.cos2x = sin 2x + cos2x + 1 \\ \\ sin 2x + (y+1)cos2x = 2y-1 \\ \\ 1+ (y+1)^2 \geq (2y-1)^2 \\ \\ \Rightarrow 3y^2-6y-1 \leq 0 \\ \\ \frac{3-2\sqrt{3}}{3} \leq y \leq \frac{3+2\sqrt{3}}{3}[/laTEX]
 
T

truongduong9083

Câu 2. Điều kiện
$$\left\{ \begin{array}{l} cos2x \neq 0 \\ cos4x \neq 1 \end{array} \right.$$
 
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