hình như sai đề

Các bước biến đổi. Bạn tự tìm kết quả nhé!

1) \(\left(\sin x-\cos x\right)\left(\cos^2x+\cos x.\sin x+\sin^2x\right)+\cos^2x-\sin^2x=0\)

<=> \(\left(\sin x-\cos x\right)\left(1+\cos x.\sin x\right)+\left(\cos x-\sin x\right)\left(\cos x+\sin x\right)=0\)

<=> \(\left(\sin x-\cos x\right)\left(\cos x+1\right)\left(\sin x+1\right)=0\)

2) \(\left(\sin^3x-2\sin^5x\right)-\left(2\cos^5x-\cos^3x\right)=0\)

<=> \(\sin^3x\left(1-2\sin^2x\right)-\cos^3x\left(2\cos^2x-1\right)=0\)

<=> \(\sin^3x.\cos2x-\cos^3x.\cos2x=0\)

<=> \(\cos2x\left(\sin^3x-\cos^3x\right)=0\)

3) ĐK: x\(\ne\frac{\pi}{2}+k\pi\)

\(\cos x\left(3.\tan x+2\right)-\left(3\tan x+2\right)=0\)

<=> \(\left(\cos x-1\right)\left(3.\tan x+2\right)=0\)

c/

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\frac{k\pi}{4}\)

\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\)

\(\Leftrightarrow4sinx.cos2x+2cos2x=2\)

\(\Leftrightarrow cos2x\left(2sinx+1\right)=1\)

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

\(\Leftrightarrow4sin^3x+2sin^2x-2sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+sinx-1\right)=0\)

\(\Leftrightarrow2sin^2x+sinx-1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b/

Nhận thấy \(sinx=0\) không phải nghiệm, pt tương đương:

\(sinx.cosx.cos2x.cos4x.cos8x=\frac{1}{16}sinx\)

\(\Leftrightarrow8sin2x.cos2x.cos4x.cos8x=sinx\)

\(\Leftrightarrow4sin4x.cos4x.cos8x=sinx\)

\(\Leftrightarrow2sin8x.cos8x=sinx\)

\(\Leftrightarrow sin16x=sinx\)

\(\Rightarrow\left[{}\begin{matrix}16x=x+k2\pi\\16x=\pi-x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{15}\\x=\frac{\pi}{17}+\frac{k2\pi}{17}\end{matrix}\right.\)

bạn xem mình giải đc chưa nhé^^

ta có: sin²x + cos²x = 1

=> sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x. cos²x

= 1 - 2sin²x. cos²x

=> sin⁸x +cos⁸x = ( sin⁴x + cos⁴x)² - 2sin⁴x. cos⁴x

= 1+ 2sin⁴x. cos⁴x - 4sin²x. cos²x

= 1+ 2sin²x. cos²x. (sin²x. cos²x -2)

= 1+ \(\dfrac{sin^22x}{2}\). (\(\dfrac{sin^22x}{4}-2\))

-Đặt t = sin²2x (đk: 0< t<1 )

=>phương trình đã cho có dạng:

1+ \(\dfrac{t}{2}.\left(\dfrac{t}{4}-2\right)\)=\(\dfrac{1}{8}\)

<=> 8 + t² -8t = 1

<=> t² - 8t +7=0

=> (t- 1)(t- 7)= 0

=>\(\left[{}\begin{matrix}t=1\left(tm\right)\\t=7\left(l\right)\end{matrix}\right.\)

Với t=1, trở lại cách đặt được sin²2x =1

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