Giúp mình với, tìm GTNN-GTLN của các hàm số

Đáp án+Giải thích các bước giải:

$\bullet$ a) $y=-2Sin3x+1$

Có: $-1\le Sin3x \le 1$

`<=>` $2\ge -2Sin3x \ge-2$

`<=>` $3\ge -2Sin3x+1 \ge-1$ ($\forall x \in \mathbb{R}$)

Vậy $Max_y=3, Min_y=-1$

$\bullet$ b) $y=3+Cos\bigg(3x+\dfrac{\pi}{4}\bigg)$

Có: $-1\le Cos\bigg(3x+\dfrac{\pi}{4}\bigg) \le 1$

`<=>` $2\le 3+Cos\bigg(3x+\dfrac{\pi}{4}\bigg) \le 4$($\forall x \in \mathbb{R}$)

Vậy $Max_y=4, Min_y=2$

$\bullet$ c) $y=Cosx+Cos\bigg(x-\dfrac{\pi}{3}\bigg)$

`<=>` $y=2.Cos\bigg(x-\dfrac{\pi}{6}\bigg).Cos\bigg(\dfrac{\pi}{6}\bigg)$

`<=>` $y=2.Cos\bigg(x-\dfrac{\pi}{6}\bigg).\dfrac{\sqrt{3}}{2}$

`<=>` $y=\sqrt{3}.Cos\bigg(x-\dfrac{\pi}{6}\bigg)$

Có $-1\le Cos\bigg(x-\dfrac{\pi}{6}\bigg) \le 1$

`<=>` $-\sqrt{3}\le \sqrt{3}.Cos\bigg(x-\dfrac{\pi}{6}\bigg) \le \sqrt{3}$($\forall x \in \mathbb{R}$)

Vậy $Max_y=\sqrt{3}, Min_y=-\sqrt{3}$

$\bullet$ d) $y=3Sin^2x+Cos2x$

`<=>` $y=3.\dfrac{1-Cos2x}{2}+Cos2x$

`<=>` $y=\dfrac{3-3Cos2x}{2}+Cos2x$

`<=>` $y=\dfrac{3-3Cos2x+2Cos2x}{2}$

`<=>` $y=\dfrac{3-Cos2x}{2}$

Có: $-1\le Cos2x \le 1$

`<=>` $1\ge -Cos2x \ge-1$

`<=>` $4\ge 3-Cos2x \ge2$

`<=>` $2\ge \dfrac{3-Cos2x}{2} \ge1$($\forall x \in \mathbb{R}$)

Vậy $Max_y=2, Min_y=1$

$\bullet$ e) $y=Sin^4x+Cos^4x$

`<=>` $y=(Sin^4x+2Sin^2x.Cos^2x+Cos^4x)-2Sin^2x.Cos^2x$

`<=>` $y=(Sin^2x+Cos^2x)^2-2Sin^2x.Cos^2x$

`<=>` $y=(1)^2-2Sin^2x.Cos^2x$

`<=>` $y=1-\dfrac{1}{2}.(2SinxCosx)^2$

`<=>` $y=1-\dfrac{1}{2}.Sin^22x$

Có: $0\le Sin^22x \le 1$

`<=>` $0\ge -\dfrac{1}{2}.Sin^22x \ge \dfrac{-1}{2}$

`<=>` $1\ge 1-\dfrac{1}{2}.Sin^22x \ge \dfrac{1}{2}$($\forall x \in \mathbb{R}$)

Vậy $Max_y=1, Min_y=\dfrac{1}{2}$