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\(\cos5x=-\sin4x\)
<=> \(\cos5x=\cos\left(4x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}5x=4x+\frac{\pi}{2}+k2\pi\\5x=-4x-\frac{\pi}{2}+k2\pi\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{\pi}{2}+k2\pi\\x=-\frac{\pi}{18}+\frac{k2\pi}{9}\end{cases}}\)
Nghiệm âm lớn nhất: \(-\frac{\pi}{18}\)
Nghiệm dương nhỏ nhất: \(\frac{\pi}{2}\)
pt <=> \(\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(2x-\frac{\pi}{3}+\frac{\pi}{2}\right)\)
<=> \(\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(2x+\frac{\pi}{6}\right)\)
<=> \(\orbr{\begin{cases}5x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\5x+\frac{\pi}{3}=\pi-2x-\frac{\pi}{6}+k2\pi\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{\pi}{14}+\frac{k2\pi}{7}\end{cases}}\)
Trên \(\left[0,\pi\right]\)có các nghiệm:
\(\frac{11\pi}{18},\frac{\pi}{14},\frac{5\pi}{14},\frac{9\pi}{14},\frac{13\pi}{14}\)
tính tổng:...
ĐKXĐ:
\(\Leftrightarrow\frac{1}{cosx}+\frac{1}{sin2x}=\frac{1}{sin2x.cos2x}\)
\(\Leftrightarrow\frac{2sinx.cos2x}{sin2x.cos2x}+\frac{cos2x}{sin2x.cos2x}=\frac{1}{sin2x.cos2x}\)
\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cos2x-sinx\right)=0\)
\(\Leftrightarrow cos2x-sinx=0\)
\(\Leftrightarrow-2sin^2x-sinx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}\\x=\frac{5\pi}{6}\end{matrix}\right.\) \(\Rightarrow\sum x=\pi\)
a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
a/
\(\Leftrightarrow sin2x\left(1+\sqrt{2}sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}sinx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sinx=-\frac{\sqrt{2}}{2}=sin\left(-\frac{\pi}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow2sin2x.cos2x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x=-sinx=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=-x+k2\pi\\4x=\pi+x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{5}\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
e/
\(sin\left(\frac{3\pi}{2}-sinx\right)=1\)
\(\Leftrightarrow\frac{3\pi}{2}-sinx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sinx=\pi+k2\pi\)
Mà \(-1\le sinx\le1\Rightarrow-1\le\pi+k2\pi\le1\)
\(\Rightarrow\) Không tồn tại k nguyên thỏa mãn
Pt đã cho vô nghiệm
f/
\(cos^2x-sin^2x+sin4x=0\)
\(\Leftrightarrow cos2x+2sin2x.cos2x=0\)
\(\Leftrightarrow cos2x\left(1+2sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
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/
\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)
\(\Leftrightarrow sin2x+cos2x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
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: ...
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)
\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)
\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)
1.
\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}\left(cos^4\frac{x}{2}-sin^4\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sinx\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sinx.cosx=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow\frac{1}{2}sin2x=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sin2x=\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
3.
ĐKXĐ: ...
\(\frac{1}{cosx}+\frac{1}{2sinx.cosx}=\frac{1}{2sinx.cosx.cos2x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cos2x-sinx\right)=0\)
\(\Leftrightarrow cos2x-sinx=0\)
\(\Leftrightarrow1-2sin^2x-sinx=0\)
\(\Leftrightarrow2sin^2x+sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
a)\(sin4x=sin\frac{\pi}{5}\)
\(\Leftrightarrow4x=\frac{\pi}{5}+k2\pi\) hoặc \(4x=\frac{4\pi}{5}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{20}+\frac{k\pi}{2}\) hoặc \(x=\frac{\pi}{5}+\frac{k\pi}{2}\)
d)cos(x+pi/18)=2/5
x+pi/18=± arccos(2/5)+k2pi
x=±arccos(2/5)-pi/18+k2pi
7.
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow8cosx=\frac{\sqrt{3}cosx+sinx}{sinx.cosx}\)
\(\Leftrightarrow8cosx.sinx.cosx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow4sin2x.cosx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow2sin3x+2sinx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow2sin3x=\sqrt{3}cosx-sinx\)
\(\Leftrightarrow sin3x=\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx\)
\(\Leftrightarrow sin\left(-3x\right)=sin\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=x-\frac{\pi}{3}+k2\pi\\-3x=\frac{4\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{2\pi}{3}+k\pi\end{matrix}\right.\)
5.
\(sin\left(2x+\frac{\pi}{2}+2\pi\right)-2cos\left(x+\frac{\pi}{2}-4\pi\right)=1+2sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)-2cos\left(x+\frac{\pi}{2}\right)=1+2sinx\)
\(\Leftrightarrow cos2x+2sinx=1+2sinx\)
\(\Leftrightarrow cos2x=1\)
\(\Rightarrow x=k\pi\)
6.
\(sin^22x-cos^28x=sin\left(10x+\frac{\pi}{2}+8\pi\right)\)
\(\Leftrightarrow\frac{1-cos4x}{2}-\frac{1+cos16x}{2}=sin\left(10x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow-\left(cos4x+cos16x\right)=2cos10x\)
\(\Leftrightarrow-2cos10x.cos6x=2cos10x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos10x=0\\cos6x=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}10x=\frac{\pi}{2}+k\pi\\6x=\pi+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{6}+\frac{k\pi}{3}\end{matrix}\right.\)