\(\text{sin2x + \frac{1}{2}tan x = \frac{3}{2} − cos 2x.}\)
Những câu hỏi liên quan
giải phương trình
a, \(2\sin\frac{x}{2}\left(\sin\frac{3x}{2}+\cos\frac{3x}{2}\right)=3-4\cos x\)
b, \(\frac{2\cos^2x+\sqrt{3}\sin2x+3}{2\cos^2x.\sin\left(x+\frac{\pi}{3}\right)}=3\left(\tan^2x+1\right)\)
a/ \(cosx-cos2x+sin2x-sinx=3-4cosx\)
\(\Leftrightarrow2sinx.cosx-sinx-2cos^2x+5cosx-2=0\)
\(\Leftrightarrow sinx\left(2cosx-1\right)-\left(2cosx-1\right)\left(cosx-2\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx-cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx-cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{3}\right)\ne0\end{matrix}\right.\) \(\Rightarrow...\)
\(\frac{2cos^2x+\sqrt{3}sin2x+3}{2cos^2x.sin\left(x+\frac{\pi}{3}\right)}=\frac{3}{cos^2x}\)
\(\Leftrightarrow2cos^2x+2\sqrt{3}sinx.cosx+3=3\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow2cos^2x-3\sqrt{3}cosx+3+2\sqrt{3}sinx.cosx-3sinx=0\)
\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx-\sqrt{3}\right)+\sqrt{3}sinx\left(2cosx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx+\sqrt{3}sinx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow...\)
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6 tháng 12 2016 lúc 22:44
giải các phương trình sau :a) sinleft(x-frac{2pi}{3}right)cos2x ; b) tanleft(2x+45^oright)tanleft(180^o-frac{x}{2}right)1 ; c) cos2x-sin^2x0 ; d) 5tan x-2cot x3 ; e) sin2x+sin^2xfrac{1}{2} ; f) sin^2frac{x}{2}+sin x-2cos^2frac{x}{2}frac{1}{2} ; g) frac{1+cos2x}{cos x}frac{sin2x}{1-cos2x}
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giải các phương trình sau :
a) \(\sin\left(x-\frac{2\pi}{3}\right)=\cos2x\) ; b) \(\tan\left(2x+45^o\right)\tan\left(180^o-\frac{x}{2}\right)=1\) ; c) \(\cos2x-\sin^2x=0\) ; d) \(5\tan x-2\cot x=3\) ; e)
\(\sin2x+\sin^2x=\frac{1}{2}\) ; f) \(\sin^2\frac{x}{2}+\sin x-2\cos^2\frac{x}{2}=\frac{1}{2}\) ; g) \(\frac{1+\cos2x}{\cos x}=\frac{\sin2x}{1-\cos2x}\)
mai đăng lại bài này nhé t làm cho h đi ngủ
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14 tháng 6 2021 lúc 9:28
Chứng minh:
a) \(tan(\frac\pi4+\frac{x}2).\frac{1+cos(\frac\pi2+x)}{sin(\frac\pi2+x)}=1\)
b) \(tan(\frac\pi4+x)=\frac{1+sin2x}{cos2x}\)
c) \(\frac{cosx}{1-sinx}=cot(\frac\pi4-\frac{x}{2})\)
d) \(tanx.tan3x=\frac{tan^22x-tan^2x}{1-tan^2x.tan^22x}\)
Chứng minh|
a) \(\frac{1+sin2x}{sinx+cosx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=sinx\)
b) \(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
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Giải phương trình:
1, \(2\tan^2x-3\tan x+2\cot^2x-3\cot x+2=0\)
2, \(\cos^23x.\cos2x-\cos^2x=0\)
3, \(\cos^22x-2\left(\cos x+\sin x\right)^2-3\sin2x+1=0\)
4, \(1-\frac{1}{\tan x}=\frac{1}{2\tan2x}\)
Mọi người giúp mình với ạ!!! Mình cảm ơn nhiều!!!
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 \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
Cho tan a=2.Tính \(B=\frac{cos^2x+sin2x+1}{2sin^2x+cos^2x+2}\)
Giải các phương trình :
1) frac{sin^4x+cos^4x}{sin2x}frac{1}{2}left(tan x+cot2xright)
2) frac{1}{sin x}+frac{1}{sinleft(x-frac{3pi}{2}right)}4sinleft(frac{7pi}{4}-xright)
3)2left(cos^42x-sin^42xright)+cos8x-cos4x0
4)frac{sin^4x+cos^4x}{5sin2x}frac{1}{2}cot2x-frac{1}{8sin2x}
5)sin^4x+cos^4x-3sin2x+frac{5}{2}sin^22x0
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Giải các phương trình :
1) \(\frac{\sin^4x+\cos^4x}{\sin2x}=\frac{1}{2}\left(\tan x+\cot2x\right)\)
2) \(\frac{1}{\sin x}+\frac{1}{\sin\left(x-\frac{3\pi}{2}\right)}=4\sin\left(\frac{7\pi}{4}-x\right)\)
3)\(2\left(\cos^42x-\sin^42x\right)+\cos8x-\cos4x=0\)
4)\(\frac{\sin^4x+\cos^4x}{5\sin2x}=\frac{1}{2}\cot2x-\frac{1}{8\sin2x}\)
5)\(\sin^4x+\cos^4x-3\sin2x+\frac{5}{2}\sin^22x=0\)
Mọi người giúp em giải bài này ạ, em cảm ơn
Bài 1: Rút gọn biểu thức:
Afrac{sin2x+sin x}{1+cos2x+cos x}
Bcotaleft(frac{1+sin^2a}{cos a}-cosaright)
Cfrac{1+cos x+cos2x+cos3x}{2cos^2x+cos x-1}
Dfrac{2cosleft(frac{pi}{2}-xright)cdotsinleft(frac{pi}{2}+xright)cdottanleft(pi-xright)}{cotleft(frac{pi}{2}+xright)cdotsinleft(pi-xright)}-2cos x
Ecos^2xcdotcot^2x+3cos^2x-cot^2x+2sin^2x
Ffrac{sin^2x+sin^2xtan^2x}{cos^2x+cos^2xtan^2x}
Gfrac{1+cos2a-cosa}{2sina-sina}
Hsin^{^{ }4}left(frac{pi}{2}+alp...
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Mọi người giúp em giải bài này ạ, em cảm ơn
Bài 1: Rút gọn biểu thức:
A=\(\frac{\sin2x+\sin x}{1+\cos2x+\cos x}\)
B=\(cota\left(\frac{1+\sin^2a}{\cos a}-cosa\right)\)
C=\(\frac{1+\cos x+\cos2x+\cos3x}{2\cos^2x+\cos x-1}\)
D=\(\frac{2\cos\left(\frac{\pi}{2}-x\right)\cdot\sin\left(\frac{\pi}{2}+x\right)\cdot\tan\left(\pi-x\right)}{\cot\left(\frac{\pi}{2}+x\right)\cdot\sin\left(\pi-x\right)}-2\cos x\)
E=\(\cos^2x\cdot\cot^2x+3\cos^2x-\cot^2x+2\sin^2x\)
\(F=\frac{\sin^2x+\sin^2x\tan^2x}{\cos^2x+\cos^2x\tan^2x}\)
\(G=\frac{1+cos2a-cosa}{2sina-sina}\)
H=\(sin^{^{ }4}\left(\frac{\pi}{2}+\alpha\right)-cos^4\left(\frac{3\pi}{2}-\alpha\right)+1\)
Bài 2: chứng minh
a) cho \(\Delta ABCchứngminhsin\frac{A+B}{2}=cos\frac{C}{2}\)
b) chứng minh biểu thức sau độc lập với biến x:
A=\(cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)\)
c)cho \(\Delta\) ABC chứng minh : sin A+sin B+ sin C= \(4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
d)CMR: \(\frac{cos2a}{1+sin2a}=\frac{cosa-sina}{cosa+sina}\)
e) CMR:\(E=\frac{sin\alpha+cos\alpha}{cos^3\alpha}=1+tan\alpha+tan^2\alpha+tan^3\alpha\)
f) CMR \(\Delta\)ABC cân khi và chỉ khi \(sinB=2cosAsinC\)
g) CM: \(\frac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
h)CM: \(\left(cos3x-cosx\right)^2+\left(sin3x-sinx\right)^2=4sin^2x\)
k) CMR trong tam giac ABC ta có: \(sin2A+sin2B+sin2C=4sinA\cdot sinB\cdot sinC\)
Bài 3: giải bất phương trình:
a)\(\frac{\left(1-3x\right)\left(2x^2+1\right)}{-2x^2-3x+5}>0\)
b)\(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\ge0\)
c)\(\frac{\left(3x-2\right)\left(x^2-9\right)}{x^2-4x+4}\le0\)
d)\(\frac{\left(2x^2+3x\right)\left(3-2x\right)}{1-x^2}\ge0\)
e)\(\frac{\left(x^2+2x+1\right)\left(x-1\right)}{3-x^2}\)
f)\(\frac{2x+1}{-x^2+x+6}\ge0\)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
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\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
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Bài 2:
\(sin\frac{A+B}{2}=sin\left(\frac{180^0-C}{2}\right)=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)
b/
\(A=cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)=cosx+2cos\left(x+\pi\right).cos\frac{\pi}{3}\)
\(=cosx-2cosx.\frac{1}{2}=0\)
c/
\(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
d/ \(\frac{cos2a}{1+sin2a}=\frac{cos^2a-sin^2a}{cos^2a+sin^2a+2sina.cosa}=\frac{\left(cosa-sina\right)\left(cosa+sina\right)}{\left(cosa+sina\right)^2}=\frac{cosa-sina}{cosa+sina}\)
e/
\(E=\frac{sina+cosa}{cos^3a}=\frac{1}{cos^2a}\left(tana+1\right)=\left(1+tan^2a\right)\left(tana+1\right)\)
\(E=tan^3a+tan^2a+tana+1\)
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