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NV
13 tháng 7 2020

\(cos11x.cos3x=cos17x.cos9x\)

\(\Leftrightarrow\frac{1}{2}\left(cos24x+cos8x\right)=\frac{1}{2}\left(cos26x+cos8x\right)\)

\(\Leftrightarrow cos24x=cos26x\)

\(\Rightarrow\left[{}\begin{matrix}26x=24x+k2\pi\\26x=-24x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{k\pi}{25}\end{matrix}\right.\)

\(sin18x.cos13x=sin9x.cos4x\)

\(\Leftrightarrow\frac{1}{2}\left(sin31x+sin5x\right)=\frac{1}{2}\left(sin13x+sin5x\right)\)

\(\Leftrightarrow sin31x=sin13x\)

\(\Rightarrow\left[{}\begin{matrix}31x=13x+k2\pi\\31x=\pi-13x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{9}\\x=\frac{\pi}{44}+\frac{k\pi}{22}\end{matrix}\right.\)

NV
13 tháng 7 2020

\(sin2x-2sinx=0\)

\(\Leftrightarrow2sinx.cosx-2sinx=0\)

\(\Leftrightarrow sinx\left(cosx-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cosx=1\end{matrix}\right.\)

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

\(cosx.cos4x-cosx.cos2x=0\)

\(\Leftrightarrow cosx\left(cos4x-cos2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos4x=cos2x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\4x=2x+k2\pi\\4x=-2x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k\pi\\x=\frac{k\pi}{3}\end{matrix}\right.\)

NV
31 tháng 7 2020

c/

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow sinx=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

d/

\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)

\(\Leftrightarrow sin2x+cos2x=1\)

\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

NV
31 tháng 7 2020

a/ ĐKXĐ:...

\(\Leftrightarrow\frac{sinx}{cosx}-\frac{\sqrt{2}}{cosx}=1\)

\(\Leftrightarrow sinx-\sqrt{2}=cosx\)

\(\Leftrightarrow sinx-cosx=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{3\pi}{4}+k2\pi\)

b/

ĐKXĐ: ...

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)

\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)

\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)

NV
6 tháng 10 2020

a.

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

NV
6 tháng 10 2020

b.

\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)

\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

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

14 tháng 9 2019

cậu có đáp án bài này chưa cho mk xin với

31 tháng 7 2018

NV
13 tháng 7 2020

\(sin3x-sinx+sin2x=0\)

\(\Leftrightarrow2cos2x.sinx+2sinx.cosx=0\)

\(\Leftrightarrow sinx\left(cos2x+cosx\right)=0\)

\(\Leftrightarrow2sinx.cos\frac{3x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{x}{2}=0\\cos\frac{3x}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{3x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)

NV
13 tháng 7 2020

\(cosx+cos3x+cos2x+cos4x=0\)

\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)

\(\Leftrightarrow cosx\left(cos2x+cos3x\right)=0\)

\(\Leftrightarrow2cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{x}{2}=0\\cos\frac{5x}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

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

NV
3 tháng 10 2019

a/ \(\Leftrightarrow sin8x+sin2x=sin12x+sin2x\)

\(\Leftrightarrow sin12x=sin8x\)

\(\Rightarrow\left[{}\begin{matrix}12x=8x+k2\pi\\12x=\pi-8x+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)

b/ \(\sqrt{2}sinx-2\sqrt{2}cosx-2+2sinx.cosx=0\)

\(\Leftrightarrow\sqrt{2}sinx\left(\sqrt{2}cosx+1\right)-2\left(\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{2}sinx-2\right)\left(\sqrt{2}cosx+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=\sqrt{2}>1\left(l\right)\\cosx=-\frac{\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{3\pi}{4}+k2\pi\)

c/ Không là hệ quả của pt nào, chắc bạn ghi nhầm đề bài

NV
9 tháng 9 2020

\(\Leftrightarrow2sinx+cos3x+sin2x-sin4x-1=0\)

\(\Leftrightarrow2sinx-1+cos3x-2cos3x.sinx=0\)

\(\Leftrightarrow2sinx-1-cos3x\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(1-cos3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cos3x=1\end{matrix}\right.\)

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

14 tháng 9 2021

a) TH1: sinx = 1 

--> x = pi/2 + k2pi (k nguyên)

TH2: sinx = -3 (loại)

14 tháng 9 2021

b) 2cosx + cos2x = 0

<=> 2cosx + 2cos^2(x) - 1 = 0

TH1: cosx = (-1 + sqrt(3))/2

TH2: cosx = (-1 - sqrt(3))/2 (loại)