Đưa về $a \sin 4x + b \cos 4x = c$ rồi dùng ĐK $a^2 + b^2 \geqslant c^2$ để giới hạn $y$
1. $y = \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x - \dfrac12 \sin 4x$
$= 1 - 3 \sin^2 x \cos^2 x - \dfrac12 \sin 4x$
$= 1 - \dfrac34 \sin^2 2x - \dfrac12 \sin 4x$
$= 1 - \dfrac38 (1 - \cos 4x) - \dfrac12 \sin 4x$
Suy ra $y - \dfrac{5}{8} = \dfrac{3}{8} \cos 4x - \dfrac12 \sin 4x$
ĐK có nghiệm: $(y - \dfrac{5}{8})^2 \leqslant \dfrac{9}{64} + \dfrac1{4}$
$\iff (y - \dfrac{5}{8})^2 \leqslant \dfrac{25}{64}$
$\iff -\dfrac{5}{8} \leqslant y - \dfrac{5}{8} \leqslant \dfrac{5}{8}$
$\iff 0 \leqslant y \leqslant \dfrac{5}{4}$
2. $y = \dfrac{\sin x + \cos x-1}{\sin x - \cos x + 3}$
Suy ra $(\sin x - \cos x + 3)y = \sin x + \cos x - 1$
$\iff (y - 1)\sin x + (-y - 1) \cos x = -1 - 3y$
ĐK: $(y-1)^2 + (-y-1)^2 \geqslant (-1-3y)^2$
$\iff 2y^2 + 2 \geqslant 9y^2 + 6y + 1$
$\iff 7y^2 + 6y - 1 \leqslant 0$
$\iff -1 \leqslant y \leqslant \dfrac17$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
