[TEX]2/cos^8x+sin^8x=17/16cos^2(2x)[/TEX]
[TEX]\Leftrightarrow (sin^4x+cos^4x)^2-2sin^4xcos^4x = \frac{17}{16}cos^22x[/TEX]
[TEX]\Leftrightarrow [(sin^2x+cos^2x)^2 - 2sin^2xcos^2x]^2 - \frac{1}{8}sin^42x = \frac{17}{16}cos^22x[/TEX]
[TEX]\Leftrightarrow (1-\frac{1}{2}sin^22x)^2 - \frac{1}{8}sin^42x = \frac{17}{16}cos^22x[/TEX]
[TEX]\Leftrightarrow 1-sin^22x + \frac{1}{8}sin^42x = \frac{17}{16}(1-2sin^22x)[/TEX]
[TEX]\Leftrightarrow 2sin^42x + sin^22x - 1 = 0[/TEX]
Nhận
[TEX]sin^22x = \frac{1}{2}[/TEX]
[TEX]\Leftrightarrow \frac{1-cos4x}{2} = \frac{1}{2}[/TEX]
[TEX]\Leftrightarrow x = \frac{\pi}{8}+k\frac{\pi}{4} (k \in Z)[/TEX]
nhớ thanks nhé ^^~