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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
7 tháng 8 2020

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

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

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

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

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

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

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

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

\(\Leftrightarrow...\)

26 tháng 10 2021

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19 tháng 8 2019
https://i.imgur.com/KATLCup.jpg
19 tháng 8 2019
https://i.imgur.com/C3DgdmP.jpg
NV
6 tháng 9 2020

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

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

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

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

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

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

27 tháng 9 2021

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

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

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

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

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

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

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

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

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

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

3 tháng 9 2021

1.

\(3sin^22x-2sin2x.cos2x-4cos^22x=2\)

\(\Leftrightarrow-\dfrac{3}{2}\left(1-2sin^22x\right)-2sin2x.cos2x-2\left(2cos^22x-1\right)=\dfrac{5}{2}\)

\(\Leftrightarrow sin4x+\dfrac{7}{2}cos4x=-\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{53}}{2}\left(\dfrac{2}{\sqrt{53}}sin4x+\dfrac{7}{\sqrt{53}}cos4x\right)=-\dfrac{5}{2}\)

\(\Leftrightarrow sin\left(4x+arccos\dfrac{2}{\sqrt{53}}\right)=-\dfrac{5}{\sqrt{53}}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+arccos\dfrac{2}{\sqrt{53}}=arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\\4x+arccos\dfrac{2}{\sqrt{53}}=\pi-arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}+\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}-\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\end{matrix}\right.\)

3 tháng 9 2021

2.

\(2\sqrt{3}cos^2x+6sinx.cosx=3+\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}\left(2cos^2x-1\right)+6sinx.cosx=3\)

\(\Leftrightarrow\sqrt{3}cos2x+3sin2x=3\)

\(\Leftrightarrow2\sqrt{3}\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)=3\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)