Cho \(\sin x+\cos x=\frac{1}{5}\). Tính \(\left|\sin x-\cos x\right|\)
Những câu hỏi liên quan
Cho \(\cos 2x = \frac{1}{4}\).
Tính: \(A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right)\); \(B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)\)
\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)
\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) = - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B = - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) = - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = - \frac{3}{8}\end{array}\)
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giải phương trình
sin xsqrt{1+2sin x}cos2x
sinleft(frac{5x}{2}-frac{pi}{4}right)-cosleft(frac{x}{2}-frac{pi}{4}right)sqrt{2}cosfrac{3x}{2}
3sqrt{tan x+1}left(sin x+2cos xright)5left(sin x+3cos xright)
sqrt{2}left(sin x+sqrt{3}cos xright)sqrt{3}cos2x-sin2x
sin2xsin4x+2left(3sin x-4sin^2x+1right)0
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giải phương trình
\(\sin x\sqrt{1+2\sin x}=\cos2x\)
\(\sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)-\cos\left(\frac{x}{2}-\frac{\pi}{4}\right)=\sqrt{2}\cos\frac{3x}{2}\)
\(3\sqrt{\tan x+1}\left(\sin x+2\cos x\right)=5\left(\sin x+3\cos x\right)\)
\(\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos2x-\sin2x\)
\(\sin2x\sin4x+2\left(3\sin x-4\sin^2x+1\right)=0\)
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 \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
Giải phương trình:
1.cos^3x.cos3x+sin^3x.sin3xfrac{sqrt{2}}{4}
2.cos^34xcos^3x.cos3x+sin^3x.sin3x
3.cos^2x-4sin^2left(frac{x}{2}-frac{pi}{4}right)+20
4.sin^4x+sin^4left(x+frac{pi}{4}right)frac{1}{4}
5.sin^6x+cos^6xfrac{5}{6}left(sin^4x+cos^4xright)
6.sin^6x+cos^6x+frac{1}{2}sinx.cosx0
7.frac{1}{2}left(sin^4x+cos^4xright)sin^2x.cos^2x+sinx.cosx
8.sin^6x+cos^6x-3cos8x+20
9.sin^4x+cos^4x-2left(sin^6frac{x}{2}+cos^6frac{x}{2}right)+10
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Giải phương trình:
1.\(cos^3x.cos3x+sin^3x.sin3x=\frac{\sqrt{2}}{4}\)
2.\(cos^34x=cos^3x.cos3x+sin^3x.sin3x\)
3.\(cos^2x-4sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)+2=0\)
4.\(sin^4x+sin^4\left(x+\frac{\pi}{4}\right)=\frac{1}{4}\)
5.\(sin^6x+cos^6x=\frac{5}{6}\left(sin^4x+cos^4x\right)\)
6.\(sin^6x+cos^6x+\frac{1}{2}sinx.cosx=0\)
7.\(\frac{1}{2}\left(sin^4x+cos^4x\right)=sin^2x.cos^2x+sinx.cosx\)
8.\(sin^6x+cos^6x-3cos8x+2=0\)
9.\(sin^4x+cos^4x-2\left(sin^6\frac{x}{2}+cos^6\frac{x}{2}\right)+1=0\)
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
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6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
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1.
\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)
\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)
\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)
\(\Leftrightarrow4cos^32x=\sqrt{2}\)
\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)
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Chứng minh rằng: (Pls help me)
a, frac{1}{sin x}+cot xcotfrac{x}{2}
b, frac{1-cos x}{sin x}tanfrac{x}{2}
c,tanfrac{x}{2}left(frac{1}{cos x}+1right)tan x
d,frac{sin2a}{2cos aleft(1+cos aright)}tanfrac{a}{2}
e,cot x+tanfrac{x}{2}frac{1}{sin x}
f,3-4cos2x+cos4x8sin^4x
g,frac{1-cos x}{sin x}frac{sin x}{1+cos x}
h,sin x+cos xsqrt{2}sinleft(x+frac{pi}{4}right)
i,sin x-cos xsqrt{2}sinleft(x-frac{pi}{4}right)
l,cos x-sin xsqrt{2}cosleft(x+frac{pi}{4}right)
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Chứng minh rằng: (Pls help me)
a, \(\frac{1}{\sin x}+\cot x=\cot\frac{x}{2}\)
b, \(\frac{1-\cos x}{\sin x}=\tan\frac{x}{2}\)
c,\(\tan\frac{x}{2}\left(\frac{1}{\cos x}+1\right)=\tan x\)
d,\(\frac{\sin2a}{2\cos a\left(1+\cos a\right)}=\tan\frac{a}{2}\)
e,\(\cot x+\tan\frac{x}{2}=\frac{1}{\sin x}\)
f,\(3-4\cos2x+\cos4x=8\sin^4x\)
g,\(\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}\)
h,\(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\)
i,\(\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\)
l,\(\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\)
a/
\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)
b/
\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)
c/
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)
\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)
d/
\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)
e/
\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)
Các câu c, e đều sử dụng kết quả từ câu b
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f/
\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)
\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)
\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)
\(=2.\left(-2sin^2x\right)^2=8sin^4x\)
g/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
h/
\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
i/
\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
j/
\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
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cho \(\sin x+\cos x=\dfrac{1}{5}\), \(x\in\left(\dfrac{\pi}{2};\pi\right)\) . Tính B = sin^5 (x) + cos^3 (x)
Giải các PT sau:1. dfrac{left(2cos2x-1right)left(sin x-3right)}{sin x}0 2.dfrac{3left(sin x+cos xright)}{sin x-cos x}2+2cos x3.dfrac{3left(sin x+tan xright)}{tan x-sin x}-2cos x24. 1+sin x+cos x+sin2x+cos2x05. 2sin xleft(1+cos2xright)+sin2x1+2cos x
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Giải các PT sau:
1. \(\dfrac{\left(2\cos2x-1\right)\left(\sin x-3\right)}{\sin x}=0\)
2.\(\dfrac{3\left(\sin x+\cos x\right)}{\sin x-\cos x}=2+2\cos x\)
3.\(\dfrac{3\left(\sin x+\tan x\right)}{\tan x-\sin x}-2\cos x=2\)
4. \(1+\sin x+\cos x+\sin2x+\cos2x=0\)
5. \(2\sin x\left(1+\cos2x\right)+\sin2x=1+2\cos x\)
1.
ĐKXĐ: \(x\ne k\pi\)
\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
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2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.
3.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)
\(\Leftrightarrow\dfrac{3\left(1+cosx\right)}{1-cosx}+2\left(1+cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(\dfrac{3}{1-cosx}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
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4.
\(\Leftrightarrow\left(sin^2x+cos^2x+2sinx.cosx\right)+\left(sinx+cosx\right)+\left(cos^2x-sin^2x\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)^2+\left(sinx+cosx\right)+\left(sinx+cosx\right)\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+cosx+1+cosx-sinx\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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giải pt : \(\frac{\cos^2x\left(\cos x-1\right)}{\sin x+\cos x}=2\left(1+\sin x\right)\)
ĐKXĐ: \(x\ne-\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow\frac{\left(1-sin^2x\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1+sinx\right)\)
\(\Leftrightarrow\frac{\left(1+sinx\right)\left(1-sinx\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1+sinx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\\\frac{\left(1-sinx\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cosx-1-sinx.cosx+sinx=2sinx+2cosx\)
\(\Leftrightarrow sinx+cosx+sinx.cosx+1=0\)
\(\Leftrightarrow\left(sinx+1\right)\left(cosx+1\right)=0\)
\(\Leftrightarrow...\)
a/sin3x+cos2x1+2sin xcos2xb/sin^3x+cos^3x2left(sin^5x+cos^5xright)c/dfrac{tan x}{sin x}-dfrac{sin x}{cos x}dfrac{sqrt{2}}{2}d/dfrac{cos xleft(cos x+2sin xright)+3sin xleft(sin x+sqrt{2}right)}{sin2x-1}1e/sin^2x+sin^23x-2cos^22x0f/dfrac{tan x-sin x}{sin^3x}dfrac{1}{cos x}g/sin2xleft(cos x+tan2xright)4cos^2xh/sin^2x+sin^23xcos^2x+cos^23xk/4sin2xdfrac{cos^2x-sin^2x}{cos^6x+sin^6x}mọi người giải giúp em với em đang cần gấp ạ
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a/\(\sin3x+\cos2x=1+2\sin x\cos2x\)
b/\(\sin^3x+\cos^3x=2\left(\sin^5x+\cos^5x\right)\)
c/\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cos x}=\dfrac{\sqrt{2}}{2}\)
d/\(\dfrac{\cos x\left(\cos x+2\sin x\right)+3\sin x\left(\sin x+\sqrt{2}\right)}{\sin2x-1}=1\)
e/\(\sin^2x+\sin^23x-2\cos^22x=0\)
f/\(\dfrac{\tan x-\sin x}{\sin^3x}=\dfrac{1}{\cos x}\)
g/\(\sin2x\left(\cos x+\tan2x\right)=4\cos^2x\)
h/\(\sin^2x+\sin^23x=\cos^2x+\cos^23x\)
k/\(4\sin2x=\dfrac{\cos^2x-\sin^2x}{\cos^6x+\sin^6x}\)
mọi người giải giúp em với em đang cần gấp ạ
1) frac{1}{cos x}+frac{1}{sin2x}frac{2}{sin4x}
2) cos3xcdottan5xsin7x
3) tan5xcdottan2x1
4) 4cos x-2cos2x-cos4x1
5) sinleft(2x+frac{5pi}{2}right)-2cosleft(x-frac{7pi}{2}right)1+2sin x
6) sin^22x-cos^28xsinleft(frac{17pi}{2}+10xright)
7) 8cos xfrac{sqrt{3}}{sin x}+frac{1}{cos x}
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1) \(\frac{1}{\cos x}+\frac{1}{\sin2x}=\frac{2}{\sin4x}\)
2) \(\cos3x\cdot\tan5x=\sin7x\)
3) \(\tan5x\cdot\tan2x=1\)
4) \(4\cos x-2\cos2x-\cos4x=1\)
5) \(\sin\left(2x+\frac{5\pi}{2}\right)-2\cos\left(x-\frac{7\pi}{2}\right)=1+2\sin x\)
6) \(\sin^22x-\cos^28x=\sin\left(\frac{17\pi}{2}+10x\right)\)
7) \(8\cos x=\frac{\sqrt{3}}{\sin x}+\frac{1}{\cos x}\)
1.
DKXĐ: \(sin4x\ne0\)
\(\Leftrightarrow\frac{4sinx.cos2x}{sin4x}+\frac{2cos2x}{sin4x}=\frac{2}{sin4x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx\left(1-2sin^2x\right)+1-2sin^2x=1\)
\(\Leftrightarrow sinx\left(1-2sin^2x-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\-2sin^2x-sinx+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
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2.
ĐKXĐ: ...
\(\Leftrightarrow\frac{cos3x.sin5x}{cos5x}=sin7x\)
\(\Leftrightarrow cos3x.sin5x=sin7x.cos5x\)
\(\Leftrightarrow sin8x+sin2x=sin12x+sin2x\)
\(\Leftrightarrow sin8x=sin12x\)
\(\Leftrightarrow\left[{}\begin{matrix}12x=8x+k2\pi\\12x=\pi-8x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
Ở nghiệm đầu tiên loại các giá trị k lẻ do đó nghiệm của pt là:
\(\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
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3.
ĐKXĐ: ...
\(\Leftrightarrow tan5x=\frac{1}{tan2x}\)
\(\Leftrightarrow tan5x=cot2x\)
\(\Leftrightarrow tan5x=tan\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow5x=\frac{\pi}{2}-2x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{14}+\frac{k\pi}{7}\)
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