1, \(sin\left(x+\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{6}}{2}\)

⇔  \(\dfrac{\sqrt{2}}{2}sin\left(x+\dfrac{\pi}{6}\right)+\dfrac{\sqrt{2}}{2}cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

⇔ \(sin\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)

2, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx=1-\sqrt{3}\)

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

⇔ sinx . si

Giải hết dùm mik đc k câu 3 luôn

a/

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

\(\Leftrightarrow3sin^2x-sinx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{2}{3}\end{matrix}\right.\)

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

2.

\(2sin^2x+\left(1-\sqrt{3}\right)sinx-\frac{\sqrt{3}}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

3.

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{4}=\frac{\pi}{8}+k2\pi\\3x+\frac{\pi}{4}=-\frac{\pi}{8}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+\frac{k2\pi}{3}\\x=-\frac{\pi}{8}+\frac{k2\pi}{3}\end{matrix}\right.\)

e cảm ơn

b/

\(sin^23x-cos^24x=sin^25x-cos^26x\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos6x-\frac{1}{2}-\frac{1}{2}cos8x=\frac{1}{2}-\frac{1}{2}cos10x-\frac{1}{2}-\frac{1}{2}cos12x\)

\(\Leftrightarrow cos6x+cos8x=cos10x+cos12x\)

\(\Leftrightarrow2cos7x.cosx=2cos11x.cosx\)

\(\Leftrightarrow cosx\left(cos11x-cos7x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos11x=cos7x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\11x=7x+k2\pi\\11x=-7x+k2\pi\end{matrix}\right.\)

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

d/

\(\Leftrightarrow2sin8x.cosx=cos\left(\frac{\pi}{2}-2x\right)+1-1-cos\left(\frac{\pi}{2}+4x\right)\) (hạ bậc vế phải)

\(\Leftrightarrow2sin8x.cosx=sin2x+sin4x\)

\(\Leftrightarrow2sin8x.cosx=2sin3x.cosx\)

\(\Leftrightarrow cosx\left(sin8x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin8x=sin3x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=3x+k2\pi\\8x=\pi-3x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)

1: \(cos^2\left(x-\frac{\pi}{5}\right)=\sin^2\left(2x+\frac45\pi\right)\)

=>\(\left[\begin{array}{l}cos\left(x-\frac{\pi}{5}\right)=\sin\left(2x+\frac45\pi\right)=cos\left(\frac{\pi}{2}-2x-\frac45\pi\right)=cos\left(-2x-\frac{3}{10}\pi\right)\\ cos\left(x-\frac{\pi}{5}\right)=-\sin\left(2x+\frac45\pi\right)=\sin\left(-2x-\frac45\pi\right)=cos\left(\frac{\pi}{2}+2x+\frac45\pi\right)=cos\left(2x+\frac{13}{10}\pi\right)\end{array}\right.\)

TH1: \(cos\left(x-\frac{\pi}{5}\right)=cos\left(-2x-\frac{3}{10}\pi\right)\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{5}=-2x-\frac{3}{10}\pi+k2\pi\\ x-\frac{\pi}{5}=2x+\frac{3}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x=-\frac{3}{10}\pi+\frac{\pi}{5}+k2\pi=-\frac{1}{10}\pi+k2\pi\\ -x=\frac{3}{10}\pi+\frac{\pi}{5}+k2\pi=-\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{30}\pi+\frac{k2\pi}{3}\\ x=\frac12\pi-k2\pi\end{array}\right.\)

TH2: \(cos\left(x-\frac{\pi}{5}\right)=cos\left(2x+\frac{13}{10}\pi\right)\)

=>\(\left[\begin{array}{l}2x+\frac{13}{10}\pi=x-\frac{\pi}{5}+k2\pi\\ 2x+\frac{13}{10}\pi=-x+\frac{\pi}{5}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-x=-\frac{\pi}{5}-\frac{13}{10}\pi+k2\pi\\ 2x+x=\frac{\pi}{5}-\frac{13}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{15}{10}\pi+k2\pi=-\frac32\pi+k2\pi\\ 3x=-\frac{11}{10}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac32\pi+k2\pi\\ x=-\frac{11}{30}\pi+\frac{k2\pi}{3}\end{array}\right.\)

2: \(\sin3x=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)+cos3x\)

=>\(\sin3x-cos3x=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=cos\left(x-\frac{\pi}{5}\right)=\sin\left(\frac{\pi}{2}-x+\frac{\pi}{5}\right)=\sin\left(-x+\frac{7}{10}\pi\right)\)

=>\(\left[\begin{array}{l}3x-\frac{\pi}{4}=-x+\frac{7}{10}\pi+k2\pi\\ 3x-\frac{\pi}{4}=\pi+x-\frac{7}{10}\pi+k2\pi=x+\frac{3}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}4x=\frac{7}{10}\pi+\frac{\pi}{4}+k2\pi=\frac{19}{20}\pi+k2\pi\\ 2x=\frac{3}{10}\pi+\frac{\pi}{4}+k2\pi=\frac{11}{20}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{19}{80}\pi+\frac{k\pi}{2}\\ x=\frac{11}{40}\pi+k\pi\end{array}\right.\)

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

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

\(\Leftrightarrow\sqrt{3}cosx+2sin^2\dfrac{x}{2}=1\)

\(\Leftrightarrow\sqrt{3}cosx-cosx=0\Leftrightarrow cosx=0\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) ( k thuộc Z )

Vậy ... 

22.

Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)

\(3tan^2x+2tanx-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{3}\right)+k\pi\end{matrix}\right.\)

Nghiệm dương nhỏ nhất của pt là: \(x=arctan\left(\dfrac{1}{3}\right)\)

Mn ơi cứu tui

a) Cách 1: Ta có:

y' = 6sin⁵x.cosx - 6cos⁵x.sinx + 6sinx.cos³x - 6sin³x.cosx = 6sin³x.cosx(sin²x - 1) + 6sinx.cos³x(1 - cos²x) = - 6sin³x.cos³x + 6sin³x.cos³x = 0.

Vậy y' = 0 với mọi x, tức là y' không phụ thuộc vào x.

Cách 2:

y = sin⁶x + cos⁶x + 3sin²x.cos²x(sin²x + cos²x) = sin⁶x + 3sin⁴x.cos²x + 3sin²x.cos⁴x + cos⁶x = (sin²x + cos²x)³ = 1

Do đó, y' = 0.

b) Cách 1:

Áp dụng công thức tính đạo hàm của hàm số hợp

(cos²u)' = 2cosu(-sinu).u' = -u'.sin2u

Ta được

y' =[sin - sin] + [sin - sin] - 2sin2x = 2cos.sin(-2x) + 2cos.sin(-2x) - 2sin2x = sin2x + sin2x - 2sin2x = 0,

vì cos = cos = .

Vậy y' = 0 với mọi x, do đó y' không phụ thuộc vào x.

Cách 2: vì côsin của hai cung bù nhau thì đối nhau cho nên

cos² = cos² '

cos² = cos² .

Do đó

y = 2 cos² + 2cos² - 2sin²x = 1 +cos + 1 +cos - (1 - cos2x) = 1 +cos + cos + cos2x = 1 + 2cos.cos(-2x) + cos2x = 1 + 2cos2x + cos2x = 1.

Do đó y' = 0.