Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
Bình Trần Thị
Xem chi tiết
Lightning Farron
12 tháng 9 2016 lúc 17:10

a)\(pt\Leftrightarrow\frac{1-cos8x}{2}+\frac{1-cos6x}{2}=\frac{1-cos4x}{2}+\frac{1-cos2x}{2}\)

\(\Leftrightarrow cos2x+cos4x=cos6x+cos8x\)

\(\Leftrightarrow2cos3x\cdot cosx=2cos7x\cdot cosx\)

\(\Leftrightarrow2cos\left(cos3x-cos7x\right)=0\)

\(\Leftrightarrow2cosx\cdot\left(-2\right)\cdot sin5x\cdot sin\left(-2x\right)=0\)

\(\Leftrightarrow cosx\cdot sin2x\cdot sin5x=0\)

\(\Leftrightarrow sin2x\cdot sin5x=0\)(do sin2x=0 <=>2sinx*cosx=0 gồm th cosx=0 r`)

\(\Leftrightarrow\left[\begin{array}{nghiempt}sin2x=0\\sin5x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{k\pi}{2}\\x=\frac{k\pi}{5}\end{array}\right.\)\(\left(k\in Z\right)\)

Lightning Farron
12 tháng 9 2016 lúc 17:18

b)\(pt\Leftrightarrow1-cos2x+1-cos4x=1+cos6x+1+cos8x\)

\(\Leftrightarrow cos2x+cos8x+cos4x+cos6x=0\)

\(\Leftrightarrow cos10x\cdot cos6x+cos10x\cdot cos2x=0\)

\(\Leftrightarrow cos10x\left(cos6x+cos2x\right)=0\)

\(\Leftrightarrow cos10x\cdot cos8x\cdot cos4x=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}cos10x=0\\cos8x=0\\cos4x=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{16}+\frac{k\pi}{8}\\x=\frac{\pi}{8}+\frac{k\pi}{4}\end{array}\right.\)

Nguyễn Việt Lâm
7 tháng 8 2020 lúc 10:57

\(\Leftrightarrow2cos^2x-1+2cos^22x-1+2cos^23x-1+2cos^24x=0\)

\(\Leftrightarrow cos2x+cos4x+cos6x+2cos^24x=0\)

\(\Leftrightarrow2cos4x.cos2x+cos4x+2cos^24x=0\)

\(\Leftrightarrow cos4x\left(2cos2x+1+2cos4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2cos4x+2cos2x+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\left(2cos^22x-1\right)+2cos2x+1=0\)

\(\Leftrightarrow4cos^22x+2cos2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}cos2x=\frac{\sqrt{5}-1}{4}=cos\left(\frac{2\pi}{5}\right)\\cos2x=\frac{-\sqrt{5}-1}{4}=cos\left(\frac{4\pi}{5}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

Cihce
26 tháng 10 2021 lúc 10:07

undefined

Invisible
Xem chi tiết
Invisible
Xem chi tiết
byun aegi park
Xem chi tiết
Lê _Ngọc_Như_Quỳnh
19 tháng 8 2019 lúc 20:20
https://i.imgur.com/KATLCup.jpg
Lê _Ngọc_Như_Quỳnh
19 tháng 8 2019 lúc 20:25
https://i.imgur.com/C3DgdmP.jpg
Lê _Ngọc_Như_Quỳnh
19 tháng 8 2019 lúc 20:33
https://i.imgur.com/cJUYRGv.jpg
Trần Tuệ Nhi
Xem chi tiết
Nguyễn Việt Lâm
6 tháng 9 2020 lúc 21:58

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=\frac{1}{2}+\frac{1}{2}cos4x+\frac{1}{2}+\frac{1}{2}cos8x\)

\(\Leftrightarrow cos8x+cos2x+cos6x+cos4x=0\)

\(\Leftrightarrow2cos5x.cos3x+2cos5x.cosx=0\)

\(\Leftrightarrow cos5x\left(cos3x+cosx\right)=0\)

\(\Leftrightarrow2cos5x.cos2x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos5x=0\\cos2x=0\\cosx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k\pi}{5}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

Nguyễn Quốc Việt
Xem chi tiết
Hồng Phúc
27 tháng 9 2021 lúc 12:57

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

\(\Leftrightarrow2sin^23x-2cos^24x=2sin^25x-2cos^26x\)

\(\Leftrightarrow2sin^23x-1+1-2cos^24x=2sin^25x-1+1-2cos^26x\)

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

\(\Leftrightarrow cos6x-cos12x+cos8x-cos10x=0\)

\(\Leftrightarrow sin9x.sin6x+sin9x.sin4x=0\)

\(\Leftrightarrow sin9x.\left(sin6x+sin4x\right)=0\)

\(\Leftrightarrow2sin9x.sin5x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin9x=0\\sin5x=0\\cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{9}\\x=\dfrac{k\pi}{5}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)