Giải phương trình
\(tan^2x+cot^2x=\frac{4+sin4x}{sin^22x}\)
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12 tháng 7 2021 lúc 21:44
Giải các Phương trình sau
a) \(sin^4\frac{x}{2}+cos^4\frac{x}{2}=\frac{1}{2}\)
b) \(sin^6x+cos^6x=\frac{7}{16}\)
c) \(sin^6x+cos^6x=cos^22x+\frac{1}{4}\)
d) \(tanx=1-cos2x\)
e) \(tan(2x+\frac\pi3).tan(\frac\pi3-x)=1\)
f) \(tan(x-15^o).cot(x+15^o)=\frac{1}{3}\)
a.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{1}{2}\)
\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)
\(\Leftrightarrow1-sin^2x=0\)
\(\Leftrightarrow cos^2x=0\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
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b.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)
\(\Leftrightarrow16-12.sin^22x=7\)
\(\Leftrightarrow3-4sin^22x=0\)
\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
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c.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos^22x+\dfrac{1}{4}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)
\(\Leftrightarrow3-3sin^22x=4cos^22x\)
\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)
\(\Leftrightarrow3=3+cos^22x\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
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Chứng minh
a) \(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{1}{2}cot^4x\)
b) \(\frac{cos2x}{cot^2x-tan^2x}=\frac{1}{4}sin^22x\)
\(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{4sin^2x.cos^2x-4\left(1-sin^2x\right)}{1-8sin^2x-\left(1-2sin^22x\right)}=\frac{4sin^2x.cos^2x-4cos^2x}{2sin^22x-8sin^2x}\)
\(=\frac{-4cos^2x\left(1-sin^2x\right)}{8sin^2x.cos^2x-8sin^2x}=\frac{-4cos^2x.cos^2x}{-8sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)
\(\frac{cos2x}{cot^2x-tan^2x}=\frac{cos2x.sin^2x.cos^2x}{cos^4x-sin^4x}=\frac{\left(cos^2x-sin^2x\right).\left(2sinx.cosx\right)^2}{4\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)}=\frac{1}{4}sin^22x\)
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Giải phương trình sau:
a) $\tan ^2x+4\cos ^2x+7=4\tan x+8\cot x$
b) $6\sin ^2x+2\cos ^2x-2\sqrt{3}\sin 2x=14\sin \left(x-\frac{\pi }{6}\right)$
giải phương trình
\(\frac{sin^22x-4\sin^2x}{sin^22x+4\sin^2x}+1=2\tan^2x\)
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x+4sin^2x}+1=2tan^2x\)
\(\Leftrightarrow\frac{4sin^2x\left(cos^2x-1\right)}{4sin^2x\left(cos^2x+1\right)}+1=\frac{2sin^2x}{cos^2x}\)
\(\Leftrightarrow\frac{cos^2x}{cos^2x+1}=\frac{1-cos^2x}{cos^2x}\)
Đặt \(cos^2x=t\Rightarrow0< t< 1\)
\(\Rightarrow\frac{t}{t+1}=\frac{1-t}{t}\Leftrightarrow t^2=1-t^2\Leftrightarrow t^2=\frac{1}{2}\)
\(\Leftrightarrow t=\frac{\sqrt{2}}{2}\Leftrightarrow cos^2x=\frac{\sqrt{2}}{2}\)
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Giải các phương trình sau:
\(5\sin^22x-6\sin4x-2\cos^2x=0\)
\(2\sin^23x-10\sin6x-\cos^23x=-2\)
\(\sin^2x\left(\tan x+1\right)=3\sin x\left(\cos x-\sin x\right)+3\)
\(6\sin x-2\cos^3x=\frac{5\sin4x.\cos x}{2\cos2x}\)
Giải các phương trình :
a) cos^2x+cos^22x-cos^23x-cos^24x0
b) cos4xcosleft(pi+2xright)-sin2xcosleft(dfrac{pi}{2}-4xright)dfrac{sqrt{2}}{2}sin4x
c) tanleft(120^0+3xright)-tanleft(140^0-xright)2sinleft(80^0+2xright)
d) tan^2dfrac{x}{2}+sin^2dfrac{x}{2}tandfrac{x}{2}+cos^2dfrac{x}{2}+cot^2dfrac{x}{2}+sin x4
e) dfrac{sin2t+2cos^2t-1}{cot t-cot3t+sin3t-sin t}cos t
Đọc tiếp
Giải các phương trình :
a) \(\cos^2x+\cos^22x-\cos^23x-\cos^24x=0\)
b) \(\cos4x\cos\left(\pi+2x\right)-\sin2x\cos\left(\dfrac{\pi}{2}-4x\right)=\dfrac{\sqrt{2}}{2}\sin4x\)
c) \(\tan\left(120^0+3x\right)-\tan\left(140^0-x\right)=2\sin\left(80^0+2x\right)\)
d) \(\tan^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\tan\dfrac{x}{2}+\cos^2\dfrac{x}{2}+\cot^2\dfrac{x}{2}+\sin x=4\)
e) \(\dfrac{\sin2t+2\cos^2t-1}{\cot t-\cot3t+\sin3t-\sin t}=\cos t\)
a)\(\sin^22x+\sqrt{3}\sin4x+3\cos^22x=0\)
b)\(\sqrt{2}\left(\sin x+\cos x\right)=\tan x+\cot x\)
b.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\sqrt{2}\left(sinx+cosx\right)=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\)
\(\Leftrightarrow\sqrt{2}\left(sinx+cosx\right)=\dfrac{1}{sinx.cosx}\)
Đặt \(sinx+cosx=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(sinx.cosx=\dfrac{t^2-1}{2}\)
Pt trở thành:
\(\sqrt{2}t=\dfrac{2}{t^2-1}\Rightarrow t^3-t-\sqrt{2}=0\)
\(\Leftrightarrow\left(t-\sqrt{2}\right)\left(t^2+\sqrt{2}t+1\right)=0\)
\(\Leftrightarrow t=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k2\pi\)
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a.
\(\Leftrightarrow sin^22x+cos^22x+\sqrt{3}sin4x+1+cos4x=0\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-2\)
\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{\sqrt{3}}{2}sin4x=-1\)
\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{3}\right)=-1\)
\(\Leftrightarrow4x-\dfrac{\pi}{3}=\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\)
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Giải phương trình:
\(\tan^2x+\cot^2x=2\left(\cos^4x+\sin^4x\right)+\sin^22x\)
Mọi người giúp mình với ạ!! Cảm ơn mọi người nhiều!!
ĐKXĐ: ...
\(\Leftrightarrow tan^2x+cot^2x=2\left(cos^4x+sin^4x+2sin^2x.cos^2x\right)\)
\(\Leftrightarrow tan^2x+cot^2x=2\left(sin^2x+cos^2x\right)^2\)
\(\Leftrightarrow tan^2x+cot^2x=2\)
\(\Leftrightarrow\left(tanx-cotx\right)^2=0\)
\(\Leftrightarrow tanx=cotx=tan\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow x=\frac{\pi}{2}-x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
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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!!!