Nhận thấy `5^2+6^2=61`
______
`A=5sin4x-6cos4x`
`=\sqrt{61}. (5/\sqrt{61}sin4x-6/\sqrt{61}cosx)`
Vì `(5/\sqrt{61})^2+(6/\sqrt{61})^2=1`
` =>` Đặt: `cosα=5/\sqrt{61};sinα=6/\sqrt{61}`
(thỏa mãn
`|cosα|<1; |sinα|<1; cos^2α+sin^2α=1)`
$\\$
`A=\sqrt{61}.(cosαsin4x-sinαcos4x)`
`=\sqrt{61}.sin(4x-α)`
Với mọi `x` ta có:
`\qquad -1\le sin(4x-α)\le 1`
`=>-\sqrt{61}\le \sqrt{61}sin(4x-α)\le \sqrt{61}`
`=>min\ (5sin4x-6cos4x)=-\sqrt{61}`