Question

III. Phương trình bậc nhất đối với sinx và cosx:

*Giải các phương trình bậc nhất đối với sinx và cosx sau đây:

(2.1)

1) \(2sinx-2cosx=\sqrt{2}\)

2) \(cosx-\sqrt{3}sinx=1\)

3) \(\sqrt{3}sin\dfrac{x}{3}+cos\dfrac{x}{2}=\sqrt{2}\)

4) \(cosx-sinx=1\)

5) \(2cosx+2sinx=\sqrt{6}\)

6) \(sin3x+\sqrt{3}cosx=\sqrt{2}\)

7) \(3sinx-2cosx=2\)

(2.3)

1) \(\left(sinx-1\right)\left(1+cosx\right)=cos^2x\)

2) \(sin\left(\dfrac{\pi}{2}+2x\right)+\sqrt{3}sin\left(\pi-2x\right)=1\)

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

4) \(sin2x+cos2x=\sqrt{2}sin3x\)

5) \(sinx=\sqrt{2}sin5x-cosx\)

6) \(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)

7) \(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)

8) \(2sin^2x+\sqrt{3}sin2x=3\)

9) \(sin^4x+cos^4\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{4}\) 

(2.3)

1) \(\dfrac{\sqrt{3}\left(1-cos2x\right)}{2sinx}=cosx\)

2) \(cotx-tanx=\dfrac{cosx-sinx}{sinx.cosx}\)

3) \(\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}=4\)

4) \(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\)

5) \(3cosx+4sinx+\dfrac{6}{3cosx+4sinx+1}=6\)

(2.4)

a) Tìm nghiệm \(x\in\left(\dfrac{2\pi}{5};\dfrac{6\pi}{7}\right)\) của phương trình \(cos7x-\sqrt{3}sin7x+\sqrt{2}=0\)

b) Tìm nghiệm \(x\in\left(0;\pi\right)\) của phương trình \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+2cos^2\left(x-\dfrac{3\pi}{4}\right)\)

(2.5) Xác định tham số m để các phương trình sau đây có nghiệm:

a) \(mcosx-\left(m+1\right)sinx=m\)

b) \(\left(2m-1\right)sinx+\left(m-1\right)cosx=m-3\)

(2.6) Tìm GTLN, GTNN (nếu có) của các hàm số sau đây:

a) \(y=3sinx-4cosx+5\)

b) \(y=cos2x+sin2x-1\)

 

Answer 1

2.1

a.

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

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

Answer 2

b.

\(cosx-\sqrt{3}sinx=1\)

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

Answer 3

c.

\(\sqrt{3}sin\dfrac{x}{3}+cos\dfrac{x}{2}=\sqrt{2}\)

Câu này đề đúng không nhỉ? Nhìn thấy có vẻ không đúng lắm

d.

\(cosx-sinx=1\)

\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

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

Answer 4

e.

\(2cosx+2sinx=\sqrt{6}\)

\(\Leftrightarrow2\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)=\sqrt{6}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{\pi}{12}+k2\pi\end{matrix}\right.\)

Answer 5

f.

\(sin3x+\sqrt{3}cosx=\sqrt{2}\)

Lại 1 câu đề không đúng

g.

\(3sinx-2cosx=2\)

\(\Leftrightarrow\dfrac{3}{\sqrt{13}}sinx-\dfrac{2}{\sqrt{13}}cosx=\dfrac{2}{\sqrt{13}}\)

Đặt \(\dfrac{3}{\sqrt{13}}=cosa\) với \(0< a< \dfrac{\pi}{2}\Rightarrow\dfrac{2}{\sqrt{13}}=sina\)

\(\Rightarrow sinx.cosa-cosx.sina=sina\)

\(\Leftrightarrow sin\left(x-a\right)=sina\)

\(\Leftrightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=\pi-a+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)

Answer 6

2.2

a.

\(\left(sinx-1\right)\left(1+cosx\right)=cos^2x\)

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

\(\Leftrightarrow\left(sinx-1\right)\left(1+cosx\right)=\left(1-sinx\right)\left(1+sinx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-1=0\\-1-cosx=1+sinx\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx+cosx=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}< -1\left(vô-nghiệm\right)\end{matrix}\right.\)

Answer 7

b.

\(sin\left(\dfrac{\pi}{2}+2x\right)+\sqrt{3}sin\left(\pi-2x\right)=1\)

\(\Leftrightarrow cos2x+\sqrt{3}sin2x=1\)

\(\Leftrightarrow\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

Answer 8

c.

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

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

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

\(\Leftrightarrow\sqrt{2}\left(cosx-sinx\right)\left(cosx+sinx\right)=cosx+sinx\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)=0\\2cos\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)

Answer 9

d.

\(sin2x+cos2x=\sqrt{2}sin3x\)

\(\Leftrightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=\sqrt{2}sin3x\)

\(\Leftrightarrow sin3x=sin\left(2x+\dfrac{\pi}{4}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

Answer 10

e.

\(sinx=\sqrt{2}sin5x-cosx\)

\(\Leftrightarrow sinx+cosx=\sqrt{2}sin5x\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sin5x\)

\(\Leftrightarrow sin5x=sin\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=x+\dfrac{\pi}{4}+k2\pi\\5x=\dfrac{3\pi}{4}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Answer 11

f.

\(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)

\(\Leftrightarrow sin8x-\sqrt{3}cos8x=cos6x+\sqrt{3}sin6x\)

\(\Leftrightarrow\dfrac{1}{2}sin8x-\dfrac{\sqrt{3}}{2}cos8x=\dfrac{1}{2}cos6x+\dfrac{\sqrt{3}}{2}sin6x\)

\(\Leftrightarrow sin\left(8x-\dfrac{\pi}{3}\right)=sin\left(6x+\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}8x-\dfrac{\pi}{3}=6x+\dfrac{\pi}{6}+k2\pi\\8x-\dfrac{\pi}{3}=\dfrac{5\pi}{6}-6x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{7}\end{matrix}\right.\)

Answer 12

g.

\(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)

\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Answer 13

h.

\(2sin^2x+\sqrt{3}sin2x=3\)

\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x=3\)

\(\Leftrightarrow\sqrt{3}sin2x-cos2x=2\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x=1\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)

Answer 14

i.

\(sin^4x+cos^4\left(x+\dfrac{\pi}{2}\right)=\dfrac{1}{4}\)

\(\Leftrightarrow\left(\dfrac{1-cos2x}{2}\right)^2+\left[\dfrac{1+cos\left(2x+\dfrac{\pi}{2}\right)}{2}\right]^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left(\dfrac{1-cos2x}{2}\right)^2+\left(\dfrac{1-sin2x}{2}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow1-2cos2x+cos^22x+1-2sin2x+sin^22x=1\)

\(\Leftrightarrow3-2\left(sin2x+cos2x\right)=1\)

\(\Leftrightarrow sin2x+cos2x=1\)

\(\Leftrightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)

Answer 15

2.3

a.

\(\dfrac{\sqrt{3}\left(1-cos2x\right)}{2sinx}=cosx\)

ĐKXĐ: \(x\ne k\pi\)

\(\sqrt{3}\left(1-cos2x\right)=2sinx.cosx\)

\(\Leftrightarrow\sqrt{3}-\sqrt{3}cos2x=sin2x\)

\(\Leftrightarrow sin2x+\sqrt{3}cos2x=\sqrt{3}\)

\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)

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

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

Answer 16

b.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(cotx-tanx=\dfrac{cosx-sinx}{sinx.cosx}\)

\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{cosx}{sinx}=\dfrac{cosx-sinx}{sinx.cosx}\)

\(\Leftrightarrow\dfrac{sin^2x-cos^2x}{sinx.cosx}=\dfrac{cosx-sinx}{sinx.cosx}\)

\(\Rightarrow\left(sinx-cosx\right)\left(sinx+cosx\right)=-\left(sinx-cosx\right)\)

\(\Rightarrow\left[{}\begin{matrix}sinx-cosx=0\\sinx+cosx=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

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

Answer 17

c.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}=4\)

\(\Rightarrow\sqrt{3}sinx+cosx=4sinx.cosx\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=sin2x\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{6}+k2\pi\\2x=\dfrac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

Answer 18

d.

\(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\)

ĐKXĐ: \(x\ne\pi+k2\pi\)

\(2\left(1+sinx\right)=1+cosx\)

\(2sinx+2=1+cosx\)

\(\Leftrightarrow cosx-2sinx=1\)

\(\Leftrightarrow\dfrac{1}{\sqrt{5}}cosx-\dfrac{2}{\sqrt{5}}sinx=\dfrac{1}{\sqrt{5}}\)

Đặt \(\dfrac{1}{\sqrt{5}}=cosa\) với \(0< a< 90^0\)

\(\Rightarrow cosx.cosa-sinx.sina=cosa\)

\(\Leftrightarrow cos\left(x+a\right)=cosa\)

\(\Leftrightarrow\left[{}\begin{matrix}x+a=a+k2\pi\\x+a=-a+k2\pi\end{matrix}\right.\)

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

Answer 19

e.

\(3cosx+4sinx+\dfrac{6}{3cosx+4sinx+1}=6\)

Đặt \(3cosx+4sinx+1=t\)

\(\Rightarrow t-1+\dfrac{6}{t}=6\)

\(\Rightarrow t^2-7t+6=0\Rightarrow\left[{}\begin{matrix}t=1\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3cosx+4sinx+1=1\\3cosx+4sinx+1=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3cosx+4sinx=0\\3cosx+4sinx=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{5}cosx+\dfrac{4}{5}sinx=0\\\dfrac{3}{5}cosx+\dfrac{4}{5}sinx=1\end{matrix}\right.\)

Đặt \(\dfrac{3}{5}=cosa\) với \(0< a< \dfrac{\pi}{2}\)

\(\Rightarrow\left[{}\begin{matrix}cosx.cosa+sinx.sina=0\\cosx.cosa+sinx.sina=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-a\right)=0\\cos\left(x-a\right)=1\end{matrix}\right.\)

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

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

Answer 20

2.4

a.

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

\(\Leftrightarrow\dfrac{1}{2}cos7x-\dfrac{\sqrt{3}}{2}sin7x=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\7x+\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{84}+\dfrac{k2\pi}{7}\\x=-\dfrac{13\pi}{84}+\dfrac{k2\pi}{7}\end{matrix}\right.\)

Do \(\dfrac{2\pi}{5}< x< \dfrac{6\pi}{7}\Rightarrow\left[{}\begin{matrix}\dfrac{2\pi}{5}< \dfrac{5\pi}{84}+\dfrac{k2\pi}{7}< \dfrac{6\pi}{7}\\\dfrac{2\pi}{5}< -\dfrac{13\pi}{84}+\dfrac{k2\pi}{7}< \dfrac{6\pi}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}k=2\\k=\left\{2;3\right\}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{53\pi}{84}\\x=\dfrac{5\pi}{12}\\x=\dfrac{59\pi}{84}\end{matrix}\right.\)

Answer 21

2.4

b.

\(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+2cos^2\left(x-\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow2-2cosx-\sqrt{3}cos2x=1+1+cos\left(2x-\dfrac{3\pi}{2}\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)

\(\Leftrightarrow\sqrt{3}cos2x-sin2x=-2cosx\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=cos\left(\pi-x\right)\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\pi-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\\x=-\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(0< x< \pi\Rightarrow x=\left\{\dfrac{5\pi}{18};\dfrac{17\pi}{18};\dfrac{5\pi}{6}\right\}\)

Answer 22

2.5

Theo điều kiện có nghiệm của pt lượng giác bậc nhất, các pt có nghiệm khi:

a.

\(m^2+\left(m+1\right)^2\ge m^2\)

\(\Leftrightarrow\left(m+1\right)^2\ge0\) (luôn thỏa mãn)

Vậy pt đã cho có nghiệm với mọi m

b.

\(\left(2m-1\right)^2+\left(m-1\right)^2\ge\left(m-3\right)^2\)

\(\Leftrightarrow4m^2-7\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}m\ge\dfrac{\sqrt{7}}{2}\\m\le-\dfrac{\sqrt{7}}{2}\end{matrix}\right.\)

Answer 23

2.6

a.

\(y=5\left(\dfrac{3}{5}sinx-\dfrac{4}{5}cosx\right)+5\)

Đặt \(\dfrac{3}{5}=cosa\) với \(0< a< \dfrac{\pi}{2}\)

\(\Rightarrow y=5\left(sinx.cosa-cosx.sina\right)+5=5sin\left(x-a\right)+5\)

Do \(-1\le sin\left(x-a\right)\le1\)

\(\Rightarrow0\le y\le10\)

\(y_{min}=0\) khi \(sin\left(x-a\right)=-1\)

\(y_{max}=10\) khi \(sin\left(x-a\right)=1\)

b.

\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)-1\)

Do \(-1\le sin\left(2x+\dfrac{\pi}{4}\right)\le1\)

\(\Rightarrow-\sqrt{2}-1\le y\le\sqrt{2}-1\)

\(y_{max}=\sqrt{2}-1\) khi \(sin\left(2x+\dfrac{\pi}{4}\right)=1\)

\(y_{min}=-\sqrt{2}-1\) khi \(sin\left(2x+\dfrac{\pi}{4}\right)=-1\)