y=sin6x+cos6x+sin4x
$= (\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x) +\sin 4x $
$= (\sin^2 x + \cos^2 x)^2 - 3\sin^2 x \cos^2 x +\sin 4x$
$= 1 - \dfrac{3}{4}\sin^2 2x + \sin 4x = 1 - \dfrac{3}{8}(1 - \cos 4x) +\sin 4x$
$= \sin 4x + \dfrac{3}{8}\cos 4x +\dfrac{5}{3}$
Dạng $y = A\sin x + B\cos x $ thì ta luôn có $-\sqrt{A^2 +B^2} \le y \le \sqrt{A^2 +B^2}$
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