a/

\(\left(\frac{sin2x}{cos2x}-\frac{sinx}{cosx}\right)cos2x=\left(\frac{sin2x.cosx-cos2x.sinx}{cos2x.cosx}\right).cos2x\)

\(=\frac{sin\left(2x-x\right)}{cosx}=\frac{sinx}{cosx}=tanx\)

b/

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

\(=1+sin^2x+cos^2x-2sinx+2cosx-2sinx.cosx\)

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

c/

\(1+cotx+cot^2x+cot^3x=1+cotx+cot^2x\left(1+cotx\right)\)

\(=\left(1+cotx\right)\left(1+cot^2x\right)=\left(1+\frac{cosx}{sinx}\right)\left(1+\frac{cos^2x}{sin^2x}\right)=\frac{sinx+cosx}{sin^3x}\)

d/

\(\frac{cos3x}{sinx}+\frac{sin3x}{cosx}=\frac{cos3x.cosx+sin3x.sinx}{sinx.cosx}=\frac{cos\left(3x-x\right)}{\frac{1}{2}2sinx.cosx}=\frac{2cos2x}{sin2x}=2cot2x\)

1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

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

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

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

a/ \(sin3x=sin\left(2x+x\right)=sin2xcosx+cos2x.sinx\)

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

\(=3sinx-4sin^3x\)

b/

\(tan2x+\frac{1}{cos2x}=\frac{sin2x}{cos2x}+\frac{1}{cos2x}=\frac{sin2x+1}{cos2x}=\frac{2sinxcosx+sin^2x+cos^2x}{cos^2x-sin^2x}\)

\(=\frac{\left(sinx+cosx\right)^2}{\left(sinx+cosx\right)\left(cosx-sinx\right)}=\frac{sinx+cosx}{cosx-sinx}=\frac{\left(sinx+cosx\right)\left(cosx-sinx\right)}{\left(cos-sinx\right)^2}\)

\(=\frac{cos^2x-sin^2x}{cos^2x+sin^2x-2sinxcosx}=\frac{1-2sin^2x}{1-sin2x}\)

c/

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

\(=\frac{2sinxcosx+2sinxcosx}{cos2x}=\frac{4sinxcosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)

d/

\(\frac{sin2x}{1+cos2x}=\frac{2sinxcosx}{1+2cos^2x-1}=\frac{2sinxcosx}{2cos^2x}=\frac{sinx}{cosx}=tanx\)

e/

Chọn D.

Ta có

b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)

c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)

\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)

ĐKXĐ: ...

\(sin3x-cos3x+sinx+cosx=\dfrac{sin3x-cos3x+sinx+cosx}{\left(sin3x+cosx\right)\left(cos3x-sinx\right)}\)

\(\Rightarrow\left[{}\begin{matrix}sin3x-cos3x+sinx+cosx=0\left(1\right)\\\left(sin3x+cosx\right)\left(cos3x-sinx\right)=1\left(2\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow3sinx-4sin^3x-4cos^3x+3cosx+sinx+cosx=0\)

\(\Leftrightarrow sinx+cosx+sin^3x+cos^3x=0\)

\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)\left(1-sinx.cosx\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(2-sinx.cosx\right)=0\)

\(\Leftrightarrow sinx+cosx=0\) (loại)

(2) \(\Leftrightarrow sin3x.cos3x-sinx.cosx-sin3x.sinx+cos3x.cosx=1\)

\(\Leftrightarrow\dfrac{1}{2}sin6x-\dfrac{1}{2}sin2x+cos4x=1\)

\(\Leftrightarrow\dfrac{1}{2}\left(3sin2x-4sin^32x\right)-\dfrac{1}{2}sin2x+1-2sin^22x=1\)

\(\Leftrightarrow sin2x-2sin^32x-2sin^22x=0\)

\(\Leftrightarrow-sin2x\left(2sin^22x+2sin2x-1\right)=0\)

\(\Leftrightarrow...\)

ĐT \(\Leftrightarrow cosx+sinx=cosx+cosxtanx\)

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

=> ĐPCM .

1: TXĐ là D=R

Khi x∈D thì -x∈D

\(f\left(-x\right)=cos\left(-x\right)+\sin^2\left(-x\right)=cosx+\sin^2x=f\left(x\right)\)

=>f(x) là hàm số chẵn

2: TXĐ là D=R

Khi x∈D thì -x∈D

\(f\left(-x\right)=\sin\left(-x\right)+cos\left(-x\right)=-\sin x+cosx\)

=>f(-x)<>f(x) và f(-x)<>-f(x)

=>f(x) không là hàm số chẵn, không là hàm số lẻ

3: TXĐ là D=R\\(\left\lbrace\frac{\pi}{2}+k\pi\right\rbrace\)

Khi x∈D thì -x∈D

\(f\left(-x\right)=\tan\left(-x\right)+2\cdot\sin\left(-x\right)=-\tan x-2\cdot\sin x=-\left(\tan x+2\cdot\sin x\right)=-f\left(x\right)\)

=>f(x) là hàm số lẻ

4: ĐKXĐ: \(2x<>\frac{\pi}{2}+k\pi\)

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

=>TXĐ làD=R\{\(\frac{\pi}{4}+\frac{k\pi}{2}\) }

Khi x∈D thì -x∈D

\(f\left(-x\right)=\tan\left(-2x\right)-\sin\left(-3x\right)=-\tan2x+\sin3x=-f\left(x\right)\)

=>f(x) là hàm số lẻ

5: TXĐ là D=R

Khi x∈D thì -x∈D

\(f\left(-x\right)=\sin\left(-2x\right)+cos\left(-x\right)=-\sin2x+cosx\)

=>f(-x)<>f(x) và f(-x)<>-f(x)

=>f(x) không là hàm số chẵn, không là hàm số lẻ

6: TXĐ là D=R\\(\left\lbrace\frac{\pi}{2}+k\pi\right\rbrace\)

Khi x∈D thì -x∈D

\(f\left(-x\right)=cos\left(-x\right)\cdot\sin^2\left(-x\right)-\tan^2\left(-x\right)=cosx\cdot\sin^2x-\tan^2x=f\left(x\right)\)

=>f(x) là hàm số chẵn

7:

TXĐ là D=R

Khi x∈D thì -x∈D

\(f\left(x\right)=cos\left(x-\frac{\pi}{4}\right)+cos\left(x+\frac{\pi}{4}\right)\)

\(=cosx\cdot cos\left(\frac{\pi}{4}\right)+\sin x\cdot\sin\left(\frac{\pi}{4}\right)+cosx\cdot cos\left(\frac{\pi}{4}\right)-\sin x\cdot\sin\left(\frac{\pi}{4}\right)\)

\(=2\cdot cosx\cdot\frac{\sqrt2}{2}=\sqrt2\cdot cosx\)

\(f\left(-x\right)=\sqrt2\cdot cos\left(-x\right)=\sqrt2\cdot cosx\)

=>f(-x)=f(x)

=>f(x) là hàm số chẵn

8:TXĐ là D=R

Khi x∈D thì -x∈D

\(f\left(-x\right)=\frac{2+cos\left(-x\right)}{1+\sin^2\left(-x\right)}=\frac{2+cosx}{1+\sin^2x}=f\left(x\right)\)

=>f(x) là hàm số chẵn

9: TXĐ là D=R

Khi x∈D thì -x∈D

\(f\left(-x\right)=\left|2+\sin\left(-x\right)\right|+\left|2-\sin\left(-x\right)\right|\)

\(=\left|2-\sin x\right|+\left|2+\sin x\right|=f\left(x\right)\)

=>f(x) là hàm số chẵn