Tìm giá trị min max của Y=sin6x-cos6x +2020

Đáp án:

$\min y=-\dfrac1{\sqrt2}+2020$ dấu "=" xảy ra $\Leftrightarrow x=-\dfrac{\pi}{24}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$

$\max y=\dfrac1{\sqrt2}+2020$ dấu "=" xảy ra $\Leftrightarrow x=\dfrac{\pi}{8}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$.

Giải thích các bước giải:

$y=\sin 6x-\cos 6x+2020$

$=\dfrac1{\sqrt2}\sin\left({6x-\dfrac{\pi}4}\right)+2020$

Do $-1\le\sin x\le1$ $\forall x$

$\Rightarrow-1\le\sin\left({6x-\dfrac{\pi}4}\right)\le1$

$\Rightarrow -\dfrac1{\sqrt2}+2020\le\dfrac1{\sqrt2}\sin\left({6x-\dfrac{\pi}4}\right)+2020\le\dfrac1{\sqrt2}+2020$

Vậy $\min y=-\dfrac1{\sqrt2}+2020$ dấu "=" xảy ra $\Leftrightarrow\sin\left({6x-\dfrac{\pi}4}\right)=-1$

$\Leftrightarrow 6x-\dfrac{\pi}4=-\dfrac{\pi}2+k2\pi$

$\Leftrightarrow x=-\dfrac{\pi}{24}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$

$\max y=\dfrac1{\sqrt2}+2020$ dấu "=" xảy ra $\Leftrightarrow\sin\left({6x-\dfrac{\pi}4}\right)=1$

$\Leftrightarrow 6x-\dfrac{\pi}4=\dfrac{\pi}2+k2\pi$

$\Leftrightarrow x=\dfrac{\pi}{8}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$.