\(cos^2x+cos^22x+cos^23x+cos^24x=\frac{3}{2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TN
0
12 tháng 9 2016
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`)
\(\Leftrightarrow\left[\begin{array}{nghiempt}sin2x=0\\sin5x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{k\pi}{2}\\x=\frac{k\pi}{5}\end{array}\right.\)\(\left(k\in Z\right)\)
12 tháng 9 2016
b)\(pt\Leftrightarrow1-cos2x+1-cos4x=1+cos6x+1+cos8x\)
\(\Leftrightarrow cos2x+cos8x+cos4x+cos6x=0\)
\(\Leftrightarrow cos10x\cdot cos6x+cos10x\cdot cos2x=0\)
\(\Leftrightarrow cos10x\left(cos6x+cos2x\right)=0\)
\(\Leftrightarrow cos10x\cdot cos8x\cdot cos4x=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}cos10x=0\\cos8x=0\\cos4x=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{16}+\frac{k\pi}{8}\\x=\frac{\pi}{8}+\frac{k\pi}{4}\end{array}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 9 2020
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=\frac{1}{2}+\frac{1}{2}cos4x+\frac{1}{2}+\frac{1}{2}cos8x\)
\(\Leftrightarrow cos8x+cos2x+cos6x+cos4x=0\)
\(\Leftrightarrow2cos5x.cos3x+2cos5x.cosx=0\)
\(\Leftrightarrow cos5x\left(cos3x+cosx\right)=0\)
\(\Leftrightarrow2cos5x.cos2x.cosx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos5x=0\\cos2x=0\\cosx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k\pi}{5}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 8 2020
Do \(\left\{{}\begin{matrix}cos^2x\ge0\\cos^22x\ge0\\cos^23x\ge0\end{matrix}\right.\)
\(\Rightarrow cos^2x+cos^22x+cos^23x\ge0\) với mọi x
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}cosx=0\\cos2x=0\\cos3x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx=0\\2cos^2x-1=0\\cos3x=0\end{matrix}\right.\)
Pt vô nghiệm (do \(cosx=0\Rightarrow2cos^2x-1=-1\ne0\))
\(\Leftrightarrow2cos^2x-1+2cos^22x-1+2cos^23x-1+2cos^24x=0\)
\(\Leftrightarrow cos2x+cos4x+cos6x+2cos^24x=0\)
\(\Leftrightarrow2cos4x.cos2x+cos4x+2cos^24x=0\)
\(\Leftrightarrow cos4x\left(2cos2x+1+2cos4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2cos4x+2cos2x+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\left(2cos^22x-1\right)+2cos2x+1=0\)
\(\Leftrightarrow4cos^22x+2cos2x-1=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=\frac{\sqrt{5}-1}{4}=cos\left(\frac{2\pi}{5}\right)\\cos2x=\frac{-\sqrt{5}-1}{4}=cos\left(\frac{4\pi}{5}\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)