a.

\(y'=\dfrac{3}{cos^2\left(3x-\dfrac{\pi}{4}\right)}-\dfrac{2}{sin^2\left(2x-\dfrac{\pi}{3}\right)}-sin\left(x+\dfrac{\pi}{6}\right)\)

b.

\(y'=\dfrac{\dfrac{\left(2x+1\right)cosx}{2\sqrt{sinx+2}}-2\sqrt{sinx+2}}{\left(2x+1\right)^2}=\dfrac{\left(2x+1\right)cosx-4\left(sinx+2\right)}{\left(2x+1\right)^2}\)

c.

\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)

1: \(\sin\left(x+\frac{\pi}{4}\right)=\frac23\)

=>\(\left[\begin{array}{l}x+\frac{\pi}{4}=\arcsin\left(\frac23\right)+k2\pi\\ x+\frac{\pi}{4}=\pi-\arcsin\left(\frac23\right)+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\arcsin\left(\frac23\right)-\frac{\pi}{4}+k2\pi\\ x=\frac34\pi-\arcsin\left(\frac23\right)+k2\pi\end{array}\right.\)

2: \(cos2x-5\cdot\sin x-3=0\)

=>\(1-2\cdot\sin^2x-5\cdot\sin x-3=0\)

=>\(-2\cdot\sin^2x-5\cdot\sin x-2=0\)

=>\(2\cdot\sin^2x+5\cdot\sin x+2=0\)

=>(sin x+2)(2 sin x+1)=0

=>2 sin x+1=0

=>sin x=-1/2

=>\(\left[\begin{array}{l}x=-\frac{\pi}{6}+k2\pi\\ x=\pi+\frac{\pi}{6}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\)

3: \(cos3x=\sin2x\)

=>\(cos3x=cos\left(\frac{\pi}{2}-2x\right)\)

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

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

a: ĐKXĐ: \(1-\sin\left(x-\frac{\pi}{8}\right)>0\) và \(2x-\frac{\pi}{4}<>\frac{\pi}{2}+k\pi\)

=>\(\sin\left(x-\frac{\pi}{8}\right)<1\) và \(2x<>\frac34\pi+k\pi\)

=>\(\sin\left(x-\frac{\pi}{8}\right)<>1\) và \(x<>\frac38\pi+k\pi\)

=>\(x-\frac{\pi}{8}<>\frac{\pi}{2}+k2\pi\) và \(x<>\frac38\pi+k\pi\)

=>\(x<>\frac58\pi+k2\pi\) và \(x<>\frac38\pi+k\pi\)

=>TXĐ là D=R\{\(\frac58\pi+k2\pi;\frac38\pi+k\pi\) }

b: ĐKXĐ: \(\begin{cases}1-cos\left(x+\frac{\pi}{3}\right)<>0\\ x-\frac{\pi}{4}<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow\begin{cases}cos\left(x+\frac{\pi}{3}\right)<>1\\ x<>\frac34\pi+k\pi\end{cases}\)

=>\(\begin{cases}x+\frac{\pi}{3}<>k2\pi\\ x<>\frac34\pi+k\pi\end{cases}\Rightarrow\begin{cases}x<>-\frac{\pi}{3}+k2\pi\\ x<>\frac34\pi+k\pi\end{cases}\)

=>TXĐ là D=R\{\(-\frac{\pi}{3}+k2\pi;\frac34\pi+k\pi\) }

c: ĐKXĐ: cosx-cos3x<>0

=>cos3x<>cosx

=>\(\begin{cases}3x<>x+k2\pi\\ 3x<>-x+k2\pi\end{cases}\Rightarrow\begin{cases}2x<>k2\pi\\ 4x<>k2\pi\end{cases}\Rightarrow\begin{cases}x<>k\pi\\ x<>\frac{k\pi}{2}\end{cases}\)

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

=>TXĐ là D=R\{\(\frac{k\pi}{2}\) }

d: ĐKXĐ: \(\sin^2x-cos^2x<>0\)

=>\(cos^2x-\sin^2x<>0\)

=>cos 2x<>0

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

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

=>TXĐ là D=R\{\(\frac{\pi}{4}+\frac{k\pi}{2}\) }
e: ĐKXĐ: \(\begin{cases}x+\frac{\pi}{3}<>k\pi\\ 3x-\frac{\pi}{4}<>\frac{\pi}{2}+k\pi\\ 3x-\frac{\pi}{4}<>k\pi\end{cases}\)

=>\(\begin{cases}x<>-\frac{\pi}{3}+k\pi\\ 3x<>\frac34\pi+k\pi\\ 3x<>\frac{\pi}{4}+k\pi\end{cases}\Rightarrow\begin{cases}x<>-\frac{\pi}{3}+k\pi\\ x<>\frac14\pi+\frac{k\pi}{3}\\ x<>\frac{1}{12}\pi+\frac{k\pi}{3}\end{cases}\)

=>TXĐ là D=R\{\(-\frac{\pi}{3}+k\pi;\frac14\pi+\frac{k\pi}{3};\frac{1}{12}\pi+\frac{k\pi}{3}\) }

1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)

⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)

⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)

⇔ 2cos²x - 5cosx + 2 = 0

⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên

2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)

⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)

⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0 

⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)

⇒ sin⁴x + cos⁴x = 48.sin⁴x . cos⁴x

⇔ (sin²x + cos²x)² - 2sin²x. cos²x = 3 . (2sinx.cosx)⁴

⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)² = 3(2sinx.cosx)⁴

⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin⁴2x

⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)

⇔ 1 - 2sin²2x = 0

⇔ cos4x = 0

⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)

⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)

⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)

⇔ sin2x - sin²2x - (1 + cos4x) = 0

⇔ sin2x - sin²2x - 2cos²2x = 0

⇔ sin2x - 2 (cos²2x + sin²2x) + sin²2x = 0

⇔ sin²2x + sin2x - 2 = 0

⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)

⇔ sin2x = 1

⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

4, cos5x + cos2x + 2sin3x . sin2x = 0

⇔ cos5x + cos2x + cosx - cos5x = 0

⇔ cos2x + cosx = 0

⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)

⇔ \(cos\dfrac{3x}{2}=0\)

⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)

⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)

Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)

⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}

Vậy các nghiệm thỏa mãn là các phần tử của tập hợp 

\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)

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

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

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

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

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

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

e/

Đề câu này chắc chắn đúng chứ bạn?

f/

\(sin^4x+cos^4x=\frac{3}{4}\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{3}{4}\)

\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}-\frac{1}{2}sin^22x=0\)

\(\Leftrightarrow1-2sin^22x=0\)

\(\Leftrightarrow cos4x=0\)

\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)

c/

\(y=sin\left(4x-\frac{\pi}{3}\right)+sin\left(\frac{\pi}{3}\right)+5\)

\(=sin\left(4x-\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}+5\)

Do \(-1\le sin\left(4x-\frac{\pi}{3}\right)\le1\)

\(\Rightarrow4+\frac{\sqrt{3}}{2}\le y\le6+\frac{\sqrt{3}}{2}\)

d/

\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin2x+5\)

\(y=6-3sin^2x.cos^2x+3sin2x\)

\(y=-\frac{3}{4}sin^22x+3sin2x+6\)

\(y=\frac{3}{4}\left(sin2x+1\right)\left(5-sin2x\right)+\frac{9}{4}\ge\frac{9}{4}\)

\(y_{min}=\frac{9}{4}\) khi \(sin2x=-1\)

\(y=\frac{3}{4}\left(sin2x-1\right)\left(3-sin2x\right)+\frac{33}{4}\le\frac{33}{4}\)

\(y_{max}=\frac{33}{4}\) khi \(sin2x=1\)

a: \(\Leftrightarrow cos2x=\dfrac{1}{\sqrt{2}}\)

=>2x=pi/4+k2pi hoặc 2x=-pi/4+k2pi

=>x=pi/8+kpi hoặc x=-pi/8+kpi

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

=>x=pi/2-3x+k2pi hoặ x=pi/2+3x+k2pi

=>4x=pi/2+k2pi hoặc -2x=pi/2+k2pi

=>x=pi/8+kpi/2 hoặc x=-pi/4-kpi

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

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

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

=>3x+3/4pi=x+pi/3+k2pi hoặc 3x+3/4pi=-x-pi/3+k2pi

=>2x=-5/12pi+k2pi hoặc 4x=-13/12pi+k2pi

=>x=-5/24pi+kpi hoặc x=-13/48pi+kpi/2

e: \(\Leftrightarrow sinx-\sqrt{3}\cdot cosx=0\)

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

=>x-pi/3=kpi

=>x=kpi+pi/3