12.

\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\le\sqrt[]{2}\)

\(\Rightarrow M=\sqrt{2}\)

13.

Pt có nghiệm khi:

\(5^2+m^2\ge\left(m+1\right)^2\)

\(\Leftrightarrow2m\le24\)

\(\Rightarrow m\le12\)

14.

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x=k2\pi\)

15.

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=3\end{matrix}\right.\)

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

Đáp án A

16.

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

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

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

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

\(\left[{}\begin{matrix}2\pi\le\dfrac{\pi}{3}+k2\pi\le2018\pi\\2\pi\le\pi+k2\pi\le2018\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1\le k\le1008\\1\le k\le1008\end{matrix}\right.\)

Có \(1008+1008=2016\) nghiệm

ĐK: `x \ne kπ`

`cot(x-π/4)+cot(π/2-x)=0`

`<=>cot(x-π/4)=-cot(π/2-x)`

`<=>cot(x-π/4)=cot(x-π/2)`

`<=> x-π/4=x-π/2+kπ`

`<=>0x=-π/4+kπ` (VN)

Vậy PTVN.

hahihihihi

Câu nào bạn, nếu mà cả thì đăng tách ra đi :)

Ok bạn =))

1.

\(sin^2x-4sinx.cosx+3cos^2x=0\)

\(\Rightarrow\dfrac{sin^2x}{cos^2x}-\dfrac{4sinx}{cosx}+\dfrac{3cos^2x}{cos^2x}=0\)

\(\Rightarrow tan^2x-4tanx+3=0\)

2.

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

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

3.

\(\Leftrightarrow2^2+m^2\ge1\)

\(\Leftrightarrow m^2\ge-3\) (luôn đúng)

Pt có nghiệm với mọi m (đề bài sai)

4.

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

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

\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)

6.

ĐKXĐ: \(cosx\ne0\)

Nhân 2 vế với \(cos^2x\)

\(sin^2x-4cosx+5cos^2x=0\)

\(\Leftrightarrow1-cos^2x-4cosx+5cos^2x=0\)

\(\Leftrightarrow\left(2cosx-1\right)^2=0\)

\(\Leftrightarrow cosx=\dfrac{1}{2}\Rightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)

6.

\(cos^2x+\sqrt{3}sinx.cosx-1=0\)

\(\Leftrightarrow-sin^2x+\sqrt{3}sinx.cosx=0\)

\(\Leftrightarrow sinx\left(sinx-\sqrt{3}cosx\right)=0\)

\(\Leftrightarrow sinx\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)=0\)

\(\Leftrightarrow sinx.sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sin\left(x-\dfrac{\pi}{3}\right)=0\end{matrix}\right.\)

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

7.

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

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

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

\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)

Tất cả \(k\in Z\)

1.

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

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

\(\Leftrightarrow x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)

Đáp án trong đề bị sai

b.

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

\(\Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}7x=\dfrac{5\pi}{12}+k2\pi\\7x=-\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

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

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

\(\Rightarrow\left[{}\begin{matrix}\dfrac{143}{120}\le k\le\dfrac{67}{24}\\\dfrac{233}{120}\le k\le\dfrac{85}{24}\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}k=1\\k=\left\{2;3\right\}\end{matrix}\right.\)

\(\Rightarrow x=\left\{\dfrac{53\pi}{84};\dfrac{5\pi}{12};\dfrac{59\pi}{84}\right\}\) 

Câu 5:

1: cos3x-sin 3x=-1

=>\(\sin3x-cos3x=1\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=1\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}3x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\ 3x-\frac{\pi}{4}=\pi-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=\frac{\pi}{2}+k2\pi\\ 3x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\ x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{array}\right.\)

2: \(\sqrt3\cdot\sin\left(\frac{x}{2}\right)-cos\left(\frac{x}{2}\right)-\sqrt2=0\)

=>\(\sqrt3\cdot\sin\left(\frac{x}{2}\right)-cos\left(\frac{x}{2}\right)=\sqrt2\)

=>\(\frac{\sqrt3}{2}\cdot\sin\left(\frac{x}{2}\right)-\frac12\cdot cos\left(\frac{x}{2}\right)=\frac{\sqrt2}{2}\)

=>\(\sin\left(\frac{x}{2}-\frac{\pi}{6}\right)=\sin\left(\frac{\pi}{4}\right)\)

=>\(\left[\begin{array}{l}\frac{x}{2}-\frac{\pi}{6}=\frac{\pi}{4}+k2\pi\\ \frac{x}{2}-\frac{\pi}{6}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{6}+\frac{\pi}{4}+k2\pi=\frac{5}{12}\pi+k2\pi\\ \frac{x}{2}=\frac34\pi+\frac{\pi}{6}+k2\pi=\frac{11}{12}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac56\pi+k4\pi\\ x=\frac{11}{6}\pi+k4\pi\end{array}\right.\)

3: 3*sin 4x+4* cos4x=5

=>\(\frac35\cdot\sin4x+\frac45\cdot cos4x=1\)

=>\(\sin\left(4x+\alpha\right)=1\)

=>\(4x+\alpha=\frac{\pi}{2}+k2\pi\)

=>\(4x=\frac{\pi}{2}-\alpha+k2\pi\)

=>\(x=\frac{\pi}{8}-\frac{\alpha}{4}+\frac{k\pi}{2}\)

Bài 4:

1: \(3\cdot\sin^23x-4\cdot\sin3x+1=0\)

=>\(3\cdot\sin^23x-3\cdot\sin3x-\sin3x+1=0\)

=>(sin 3x-1)(3sin 3x-1)=0

TH1: sin 3x-1=0

=>sin 3x=1

=>\(3x=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{6}+\frac{k2\pi}{3}\)

TH2: 3 sin 3x-1=0

=>3sin 3x=1

=>sin 3x=1/3

=>\(\left[\begin{array}{l}3x=\arcsin\left(\frac13\right)+k2\pi\\ 3x=\pi-\arcsin\left(\frac13\right)+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\cdot\arcsin\left(\frac13\right)+\frac{k2\pi}{3}\\ x=\frac{\pi}{3}-\frac13\cdot\arcsin\left(\frac13\right)+\frac{k2\pi}{3}\end{array}\right.\)

2: \(4\cdot cos^2\left(\frac{x}{2}\right)-1=0\)

=>\(4\cdot cos^2\left(\frac{x}{2}\right)=1\)

=>\(cos^2\left(\frac{x}{2}\right)=\frac14\)

=>\(\left[\begin{array}{l}cos\left(\frac{x}{2}\right)=\frac12\\ cos\left(\frac{x}{2}\right)=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{3}+k2\pi\\ \frac{x}{2}=-\frac{\pi}{3}+k2\pi\\ \frac{x}{2}=\frac23\pi+k2\pi\\ \frac{x}{2}=-\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{2\pi}{3}+k4\pi\\ x=-\frac23\pi+k4\pi\\ x=\frac43\pi+k4\pi\\ x=-\frac43\pi+k4\pi\end{array}\right.\)

3: \(3\cdot\tan^24x-\sqrt3\cdot\tan4x=0\)

=>\(\sqrt3\cdot\tan4x\left(\sqrt3\cdot\tan4x-1\right)=0\)

TH1: tan 4x=0

=>\(4x=k\pi\)

=>\(x=\frac{k\pi}{4}\)

TH2: \(\sqrt3\cdot\tan4x-1=0\)

=>\(\tan4x=\frac{1}{\sqrt3}\)

=>\(4x=\frac{\pi}{6}+k\pi\)

=>\(x=\frac{\pi}{24}+\frac{k\pi}{4}\)

5: \(\sin^2x+cosx-1=0\)

=>\(1-cos^2x+cosx-1=0\)

=>\(-cos^2x+cosx=0\)

=>cosx(cosx-1)=0

TH1: cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

TH2: cos x-1=0

=>cosx =1

=>\(x=k2\pi\)

6: \(\cot^22x-2\cdot\cot2x-3=0\)

=>(cot 2x-3)(cot 2x+1)=0

TH1: cot 2x-3=0

=>cot 2x=3

=>\(2x=arc\cot\left(3\right)+k\pi\)

=>\(x=\frac12\cdot arc\cot\left(3\right)+\frac{k\pi}{2}\)

TH2: cot 2x+1=0

=>cot 2x=-1

=>\(2x=-\frac{\pi}{4}+k\pi\)

=>\(x=-\frac{\pi}{8}+\frac{k\pi}{2}\)

1.

\(pt\Leftrightarrow sin4x\left(sin5x+sin3x\right)=sin2x.sinx\)

\(\Leftrightarrow2sin^24x.cosx=sin2x.sinx\)

\(\Leftrightarrow2sin^24x.cosx=2sin^2x.cosx\)

\(\Leftrightarrow2cosx.\left(sin^24x-sin^2x\right)=0\)

\(\Leftrightarrow2cosx.\left(sin4x-sinx\right)\left(sin4x+sinx\right)=0\)

\(\Leftrightarrow8cosx.sin\dfrac{5x}{2}.cos\dfrac{3x}{2}.sin\dfrac{5x}{2}.cos\dfrac{3x}{2}=0\)

​\(\Leftrightarrow8cosx.sin5x.sin3x=0\)​

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin5x=0\\sin3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k\pi}{5}\\x=\dfrac{k\pi}{3}\end{matrix}\right.\)

\(pt\Leftrightarrow sin8x+sin2x=sin16x+sin2x\)

\(\Leftrightarrow sin8x=2sin8x.cos8x\)

\(\Leftrightarrow sin8x\left(1-2cos8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin8x=0\\cos8x=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{8}\\x=\pm\dfrac{\pi}{24}+\dfrac{k\pi}{4}\end{matrix}\right.\)

d: Gọi (d1): ax+by+c=0 là ảnh của (d) qua phép vị tự tâm I(2;-1), tỉ số k=2

=>(d1)//(d)

=>(d1): x+2y+c=0

Lấy A(2;1) thuộc (d)

Lấy B(x;y) là ảnh của A(2;1) qua phép vị tự tâm I(2;-1), tỉ số k=2

=>\(\overrightarrow{IB}=2\cdot\overrightarrow{IA}\)

=>x-2=2(2-2)=0 và y-1=2*(1+1)=2*2=4

=>x=2 và y=5

=>B(2;5)

Thay x=2 và y=5 vào x+2y+c=0, ta được:

2+2*5+c=0

=>c+12=0

=>c=-12

=>(d1): x+2y-12=0

\(sin^2x+\sqrt{3}sinxcosx=1\)

\(\Leftrightarrow sin^2x+\sqrt{3}sinxcosx=sin^2x+cos^2x\)

\(\Leftrightarrow cosx\left(\sqrt{3}sinx-cosx\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=0\\\sqrt{3}sinx=cosx\end{cases}}\Leftrightarrow\orbr{\begin{cases}cosx=0\\tanx=\frac{1}{\sqrt{3}}\end{cases}}\)

Từ đây suy ra nghiệm.