<=> sin5x=5sinx

<=> Sin5x-sinx=4sinx

<=> 2cos3x.sin2x=4sinx

<=>4cos3x.sinx.cosx=4sinx

<=>(cos3x.cosx-1).sinx=0

Sinx=0 hoặc cos3x.cosx -1=0

TH1. Sinx=0 => x=kπ 

TH2: cos3x.cosx-1=0

<=> Cos3x.cosx=1

<=>cos4x + cos2x =2

<=> 2cos ²2x -1 +cos2x -2=0

<=> 2cos ²2x +cos 2x -3=0

Cos 2x= 1 =>. X=kπ/2

Cos2x= -3/2 <-1(loai)

Vậy x=kπ/2

ĐK: \(x\ne k\pi\)

\(\dfrac{sin5x}{5sinx}=1\)

\(\Leftrightarrow sin5x=5sinx\)

\(\Leftrightarrow sin5x-sinx=4sinx\)

\(\Leftrightarrow2cos3x.sin2x=4sinx\)

\(\Leftrightarrow4sinx.cosx.cos3x=4sinx\)

\(\Leftrightarrow cosx.cos3x=1\) (Vì \(sinx\ne0\))

\(\Leftrightarrow\dfrac{1}{2}\left(cos4x+cos2x\right)=1\)

\(\Leftrightarrow2cos^22x-1+cos2x=2\)

\(\Leftrightarrow2cos^22x+cos2x-3=0\)

\(\Leftrightarrow\left(cos2x-1\right)\left(2cos2x+3\right)=0\)

\(\Leftrightarrow cos2x=1\) (Vì \(2cos2x+3>0\))

\(\Leftrightarrow x=k\pi\left(l\right)\)

Vậy phương trình đã cho vô nghiệm

a, ta có 2x + π/3 = 3π/4 +k2π hoặc 2x + π/3 = -3π/4 + k2π

=> x= 5π/24 + kπ hoặc x= -13π/24 +kπ

b, đề sai phải ko

c,  cos²2x - sin²2x - 2sinx -1=0

<=> -2sin²2x -2sin2x =0

<=> sin2x=0 hoặc sin2x=-1

<=> x=kπ hoặc x= π/2 + kπ ; x=-π/4 +kπ hoặc x=5π/8 + kπ

d, cos5xcosπ/4 - sin5xsinπ/4 = -1/2

   cos( 5x + π/4 ) = -1/2

   <=> x=π/12 +k2π/5 hoặc x= -11π/60 + k2π/5

f,4x+π/3=3π/10 -x +k2π  hoặc 4x+π/3 = x - 3π/10 +k2π

<=> x =-π/150 + k2π/5 hoặc x = π/90 +k2π/3

Câu đầu đơn giản là ko dịch được \(cos^22\times x/2\) nghĩa là gì :)

\(sin5x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow sin5x=-\left(cos^2x-sin^2x\right)\)

\(\Leftrightarrow sin5x=-cos2x\)

\(\Leftrightarrow sin5x=sin\left(2x-\frac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=2x-\frac{\pi}{2}+k2\pi\\5x=\frac{3\pi}{2}-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(1+sin5x=2cos^2\frac{x}{2}\)

\(\Leftrightarrow sin5x=2cos^2\frac{x}{2}-1\)

\(\Leftrightarrow sin5x=cosx\)

\(\Leftrightarrow sin5x=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Leftrightarrow...\)

2cos²x/2

sin3x + 1=2sin²2x

<=> sin3x + 1 = 2\(\dfrac{1-cos4x}{2}\)

<=> sin3x + 1 = 1 - cos4x

<=> sin3x = -cos4x

<=> sin3x + cos4x = 0

<=> \(\dfrac{\sqrt{2}}{2}\)sin3x + \(\dfrac{\sqrt{2}}{2}\)cos4x = 0 (chia 2 vế cho \(\sqrt{2}\)).

<=> cos\(\dfrac{\pi}{4}\)sin3x + sin\(\dfrac{\pi}{4}\)cos4x = 0

<=> sin (3x+\(\dfrac{\pi}{4}\)) = 0

<=> sin(3x+\(\dfrac{\pi}{4}\)) = sin0

<=> \(\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=0+k2\pi\\3x+\dfrac{\pi}{4}=\pi-0+k2\pi\end{matrix}\right.\)(k\(\in\)Z)

<=>\(\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(k\(\in\)Z)

\(\Leftrightarrow sin5x+sinx-\left(1-2sin^2x\right)=0\)

\(\Leftrightarrow2sin3x.cos2x-cos2x=0\)

\(\Leftrightarrow cos2x\left(2sin3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

\(sin5x+sinx+2sin^2x=1\)

\(\Leftrightarrow\left(sin5x+sinx\right)-\left(1-2sin^2x\right)=0\)

\(\Leftrightarrow2sin3x.cos2x-cos2x=0\)

\(\Leftrightarrow cos2x\left(2sin3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\end{matrix}\right.\)

Vậy...

\(sin5x+sinx+2sin^2x=1\)

\(\Leftrightarrow2sin3x.cos2x-cos2x=0\)

\(\Leftrightarrow cos2x\left(2sin3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

Đáp án D

\(\sin\left(5x\right)+\sin\left(3x\right)+2\cos\left(x\right)=1+\sin\left(4x\right)\)

\(\Leftrightarrow2\sin\left(4x\right)\cos\left(x\right)-\sin\left(4x\right)+2\cos\left(x\right)-1=0\)

\(\Leftrightarrow\sin\left(4x\right)(2\cos\left(x\right)-1)+(2\cos\left(x\right)-1)=0\)

\(\Leftrightarrow(2\cos\left(x\right)-1)(\sin\left(4x\right)+1)=0\)

\(\Rightarrow\left[{}\begin{matrix}\cos\left(x\right)=\dfrac{1}{2}\\\sin\left(4x\right)=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{-\pi}{2}+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{-\pi}{8}+k\dfrac{\pi}{2}\end{matrix}\right.\)