a, \(y=sin^2x-2sinx+3cos^2x\)

\(=sin^2x-2sinx+3\left(1-sin^2x\right)\)

\(=3-2sinx-2sin^2x\)

Đặt \(sinx=t\left(t\in\left[0;1\right]\right)\)

\(\Rightarrow y=f\left(t\right)=3-2t-2t^2\)

\(\Rightarrow y_{min}=min\left\{f\left(0\right);f\left(1\right)\right\}=-1\)

\(y_{max}=max\left\{f\left(0\right);f\left(1\right)\right\}=3\)

b, \(y=sinx-cosx+sin2x+5\)

\(=sinx-cosx-\left(sinx-cosx\right)^2+6\)

Đặt \(sinx-cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(\Rightarrow y=f\left(t\right)=-t^2+t+6\)

\(\Rightarrow y_{min}=min\left\{f\left(-\sqrt{2}\right);f\left(0\right)\right\}=4-\sqrt{2}\)

\(y_{max}=max\left\{f\left(-\sqrt{2}\right);f\left(0\right)\right\}=6\)

c, \(y=sinx-cosx+sinx.cosx-3\)

\(=sinx-cosx-\dfrac{1}{2}\left(sinx-cosx\right)^2-\dfrac{5}{2}\)

Đặt \(sinx-cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(\Rightarrow y=f\left(t\right)=-\dfrac{1}{2}t^2+t-\dfrac{5}{2}\)

\(\Rightarrow y_{min}=min\left\{f\left(-\sqrt{2}\right);f\left(\sqrt{2}\right);f\left(1\right)\right\}=-\dfrac{7+2\sqrt{2}}{2}\)

\(y_{max}=max\left\{f\left(-\sqrt{2}\right);f\left(\sqrt{2}\right);f\left(1\right)\right\}=-2\)

a) Đk: sinx \(\ne\)0<=>x\(\ne\)k\(\Pi\)

pt<=>\(\sqrt{3}\)(1-cos2x)-cosx=0

<=>\(\sqrt{3}\)[1-(2cos²x-1)]-cosx=0

<=>2\(\sqrt{3}\)-2\(\sqrt{3}\)cos²x-cosx=0

<=>\(\left\{{}\begin{matrix}cosx=\dfrac{\sqrt{3}}{2}\\cosx=-\dfrac{2\sqrt{3}}{3}< -1\left(loai\right)\end{matrix}\right.\)

tới đây bạn tự giải cho quen, chứ chép thì thành ra không hiểu gì thì khổ

b)pt<=>2sin2x+2sin²x=1

<=>2sin2x+2sin²x=sin²x+cos²x

<=>4sinx.cosx+sin²x-cos²x=0

Tới đây là dạng của pt đẳng cấp bậc 2, ta thấy cosx=0 không phải là nghiệm của pt nên ta chia cả hai vế của pt cho cos²x:

pt trở thành:

4tanx+tan²x-1=0

<=>\(\left[{}\begin{matrix}tanx=-2+\sqrt{2}\\tanx=-2-\sqrt{5}\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=arctan\left(-2+\sqrt{5}\right)+k\Pi\\x=arctan\left(-2-\sqrt{5}\right)+k\Pi\end{matrix}\right.\)(k thuộc Z)

Chú ý: arctan tương ứng ''SHIFT tan'' (khi thử nghiệm trong máy tính)

c)Đk: cosx\(\ne\)0<=>x\(\ne\)\(\dfrac{\Pi}{2}\)+kpi

pt<=>cos²x+\(\sqrt{3}\)sin²x=1

<=>1-sin²x+\(\sqrt{3}\)sin²x-1=0

<=>(\(\sqrt{3}\)-1)sin²x=0

<=>sinx=0<=>x=k\(\Pi\)(k thuộc Z)

d)

pt<=>\(\sqrt{3}\)sin7x-cos7x=\(\sqrt{2}\)

Khúc này bạn coi SGK trang 35 người ta giả thích rõ ràng rồi

pt<=>\(\dfrac{\sqrt{3}}{2}\)sin7x-\(\dfrac{1}{2}\)cos7x=\(\dfrac{\sqrt{2}}{2}\)

<=>sin(7x-\(\dfrac{\Pi}{3}\))=\(\dfrac{\sqrt{2}}{2}\)

<=>sin(7x-\(\dfrac{\Pi}{3}\))=sin\(\dfrac{\Pi}{4}\)

Tới đây bạn tự giải nhé, giải ra nghiệm rồi kiểm tra xem nghiệm nào thuộc khoảng ( đề cho) rồi kết luận

Câu d) mình nhầm nhé

<=>sin(7x-\(\dfrac{\Pi}{6}\))=\(\dfrac{\sqrt{2}}{2}\) mới đúng sorry

3.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm

a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)

\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)

\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)

\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)

b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx+\sqrt{3}\cdot cosx=0\)

\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)

\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)

(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)

\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)

\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)

\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)