a: \(sinx=sin\left(\dfrac{\Omega}{4}\right)\)

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

b: cos2x=cosx

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

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

c:

ĐKXĐ: \(x-\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)

=>\(x< >\dfrac{5}{6}\Omega+k\Omega\)

 \(tan\left(x-\dfrac{\Omega}{3}\right)=\sqrt{3}\)

=>\(x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)

=>\(x=\dfrac{2}{3}\Omega+k\Omega\)

d:

ĐKXĐ: \(2x+\dfrac{\Omega}{6}< >k\Omega\)

=>\(2x< >-\dfrac{\Omega}{6}+k\Omega\)

=>\(x< >-\dfrac{1}{12}\Omega+\dfrac{k\Omega}{2}\)

 \(cot\left(2x+\dfrac{\Omega}{6}\right)=cot\left(\dfrac{\Omega}{4}\right)\)

=>\(2x+\dfrac{\Omega}{6}=\dfrac{\Omega}{4}+k\Omega\)

=>\(2x=\dfrac{1}{12}\Omega+k\Omega\)

=>\(x=\dfrac{1}{24}\Omega+\dfrac{k\Omega}{2}\)

`a)sin x =4/3`

`=>` Ptr vô nghiệm vì `-1 <= sin x <= 1`

`b)sin 2x=-1/2`

`<=>[(2x=-\pi/6+k2\pi),(2x=[7\pi]/6+k2\pi):}`

`<=>[(x=-\pi/12+k\pi),(x=[7\pi]/12+k\pi):}`    `(k in ZZ)`

`c)sin(x - \pi/7)=sin` `[2\pi]/7`

`<=>[(x-\pi/7=[2\pi]/7+k2\pi),(x-\pi/7=[5\pi]/7+k2\pi):}`

`<=>[(x=[3\pi]/7+k2\pi),(x=[6\pi]/7+k2\pi):}`     `(k in ZZ)`

`d)2sin (x+pi/4)=-\sqrt{3}`

`<=>sin(x+\pi/4)=-\sqrt{3}/2`

`<=>[(x+\pi/4=-\pi/3+k2\pi),(x+\pi/4=[4\pi]/3+k2\pi):}`

`<=>[(x=-[7\pi]/12+k2\pi),(x=[13\pi]/12+k2\pi):}`    `(k in ZZ)`

a: sin x=4/3

mà -1<=sinx<=1

nên \(x\in\varnothing\)

b: sin 2x=-1/2

=>2x=-pi/6+k2pi hoặc 2x=7/6pi+k2pi

=>x=-1/12pi+kpi và x=7/12pi+kpi

c: \(sin\left(x-\dfrac{pi}{7}\right)=sin\left(\dfrac{2}{7}pi\right)\)

=>x-pi/7=2/7pi+k2pi hoặc x-pi/7=6/7pi+k2pi

=>x=3/7pi+k2pi và x=pi+k2pi

d: 2*sin(x+pi/4)=-căn 3

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

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

=>x=-7/12pi+k2pi hoặc x=19/12pi+k2pi

a: \(sin\left(x-\dfrac{\Omega}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

=>\(sin\left(x-\dfrac{\Omega}{4}\right)=sin\left(-\dfrac{\Omega}{4}\right)\)

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

b: \(cos\left(x+\dfrac{\Omega}{4}\right)=cos\left(\dfrac{3}{4}\Omega\right)\)

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

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)

c: ĐKXĐ: \(\left\{{}\begin{matrix}2x< >\dfrac{\Omega}{2}+k\Omega\\x+\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\\x< >\dfrac{1}{6}\Omega+k\Omega\end{matrix}\right.\)

\(tan2x=tan\left(x+\dfrac{\Omega}{3}\right)\)

=>\(2x=x+\dfrac{\Omega}{3}+k\Omega\)

=>\(x=\dfrac{\Omega}{3}+k\Omega\)

d: ĐKXĐ: \(2x< >k\Omega\)

=>\(x< >\dfrac{k\Omega}{2}\)

\(cot2x=-\dfrac{\sqrt{3}}{3}\)

=>\(cot2x=cot\left(-\dfrac{\Omega}{3}\right)\)

=>\(2x=-\dfrac{\Omega}{3}+k\Omega\)

=>\(x=-\dfrac{\Omega}{6}+\dfrac{k\Omega}{2}\)

a: pi/2<x<pi

=>cosx<0

=>\(cosx=-\sqrt{1-\left(\dfrac{1}{5}\right)^2}=-\dfrac{2\sqrt{6}}{5}\)

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{5}\cdot\dfrac{-2\sqrt{6}}{5}=\dfrac{-4\sqrt{6}}{25}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{24}{25}-1=\dfrac{48}{25}-1=\dfrac{23}{25}\)

\(tan2x=-\dfrac{4\sqrt{6}}{25}:\dfrac{23}{25}=-\dfrac{4\sqrt{6}}{23}\)

\(cot2x=1:\dfrac{-4\sqrt{6}}{23}=\dfrac{-23}{4\sqrt{6}}\)

b: \(sin\left(x-\dfrac{pi}{6}\right)=sinx\cdot cos\left(\dfrac{pi}{6}\right)-cosx\cdot sin\left(\dfrac{pi}{6}\right)\)

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

\(=\dfrac{1}{5}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{-2\sqrt{6}}{5}\cdot\dfrac{1}{2}=\dfrac{\sqrt{3}+2\sqrt{6}}{10}\)

c: \(cos\left(x-\dfrac{pi}{3}\right)=cosx\cdot cos\left(\dfrac{pi}{3}\right)+sinx\cdot sin\left(\dfrac{pi}{3}\right)\)

\(=-\dfrac{2\sqrt{6}}{5}\cdot\dfrac{1}{2}+\dfrac{1}{5}\cdot\dfrac{1}{2}=\dfrac{-2\sqrt{6}+1}{10}\)

d: \(tan\left(x-\dfrac{pi}{4}\right)=\dfrac{tanx-tan\left(\dfrac{pi}{4}\right)}{1+tanx\cdot tan\left(\dfrac{pi}{4}\right)}\)

\(=\dfrac{tanx-1}{1+tanx}\)

\(=\dfrac{\dfrac{1}{-2\sqrt{6}}-1}{1+\dfrac{1}{-2\sqrt{6}}}=\dfrac{-25-4\sqrt{6}}{23}\)

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 

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

\(sin\left(x\right)+\left[sin\left(x+\dfrac{2\pi}{5}\right)-sin\left(x+\dfrac{\pi}{5}\right)\right]+\left[sin\left(x+\dfrac{4\pi}{5}\right)-sin\left(x+\dfrac{3\pi}{5}\right)\right]\)

\(=sin\left(x\right)+2cos\left(x+\dfrac{3\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)+2cos\left(x+\dfrac{7\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)\)

\(=sin\left(x\right)+2sin\left(\dfrac{\pi}{10}\right)\left[cos\left(x+\dfrac{3\pi}{10}\right)+cos\left(x+\dfrac{7\pi}{10}\right)\right]\)

\(=sin\left(x\right)+4sin\left(\dfrac{\pi}{10}\right)cos\left(\dfrac{\pi}{5}\right)cos\left(x+\dfrac{\pi}{2}\right)\)

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

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

\(=sin\left(x\right)-sin\left(x\right)\)

\(=0\)

a: cos3x=8

mà -1<=cos3x<=1

nên \(x\in\varnothing\)

b; \(-2\cdot cosx+\sqrt{3}=0\)

=>\(-2\cdot cosx=-\sqrt{3}\)

=>\(cosx=\dfrac{\sqrt{3}}{2}\)

=>x=pi/6+k2pi hoặc x=-pi/6+k2pi

c: cos(3x-pi/6)=0

=>3x-pi/6=pi/2+k2pi

=>3x=2/3pi+k2pi

=>x=2/9pi+k2pi/3

d: cos(x+2/3pi)=cos(pi/5)

=>x+2/3pi=pi/5+k2pi hoặc x+2/3pi=-pi/5+k2pi

=>x=-7/15pi+k2pi hoặc x=-13/15pi+k2pi

e: cos^2(3x)=4

=>cos3x=2(loại) hoặc cos3x=-2(loại)