Giải pt cot(2x+1)/6=tan1/3
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Đạo hàm
y= sin(2x+π/6) + cos3x - tan1/4x - cot√5x+1
\(y'=2cos\left(2x+\frac{\pi}{6}\right)-3sin3x-\frac{1}{4cos^2\frac{1}{4x}}+\frac{5}{2\sqrt{5x+1}.sin^2\sqrt{5x+1}}\)
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giải pt: cot ( 2x - 3 ) = 1/4
\(\Leftrightarrow2x-3=arccot\left(\dfrac{1}{4}\right)+k\pi\)
\(\Rightarrow x=\dfrac{3}{2}+\dfrac{1}{2}arccot\left(\dfrac{1}{4}\right)+\dfrac{k\pi}{2}\)
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Giải pt: 3cosx - 2= -3(1-cosx)cot^2x
Giải pt sau:
\(2\left(Tan^2x-Cot^2x\right)-5\left(Tanx+Cotx\right)+6=0\)
Giải các pt sau:
a) \(\sin\left(3x+60^o\right)=\dfrac{1}{2}\)
b) \(\cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{-\sqrt{2}}{2}\)
c) \(\tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)
d) \(\cot\left(2x+\pi\right)=-1\)
a, Ta có : \(\sin\left(3x+60\right)=\dfrac{1}{2}\)
\(\Rightarrow3x+60=30+2k180\)
\(\Rightarrow3x=2k180-30\)
\(\Leftrightarrow x=120k-10\)
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b, Ta có : \(\cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow2x-\dfrac{\pi}{3}=\dfrac{3}{4}\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{13}{24}\pi+k\pi\)
Vậy ...
c, Ta có : \(tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)
\(\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)
Vậy ...
d, Ta có : \(\cot\left(2x+\pi\right)=-1\)
\(\Rightarrow2x+\pi=\dfrac{3}{4}\pi+k\pi\)
\(\Leftrightarrow x=-\dfrac{1}{8}\pi+\dfrac{k}{2}\pi\)
Vậy ...
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a) \(sin\left(3x+60^0\right)=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{3}\right)=sin\dfrac{\pi}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))
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b) Pt\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\dfrac{3\pi}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)(\(k\in Z\))
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c) Pt \(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=tan\dfrac{\pi}{3}\)
\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi,k\in Z\)\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi,k\in Z\)
Vậy...
d) Pt \(\Leftrightarrow tan\left(2x+\pi\right)=-1\)
\(\Leftrightarrow2x+\pi=-\dfrac{\pi}{4}+k\pi,k\in Z\)
\(\Leftrightarrow x=-\dfrac{5\pi}{8}+\dfrac{k\pi}{2},k\in Z\)
Vậy...
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Mấy bạnn giải chii tiết raa giúp mik với nhaa Câu 1: nghiệm dương nhỏ nhất của pt tan x=tan (6π/5) A. x=π/5 B. x=6π/5 C. x=6/5 D. x=6π Câu 2: tìm nghiệm thuộc đoạn [0;π] của pt cot 2x=cot(π/2-x) A. 2 B. 3 C.1 D.4 Câu 3: tìm tổng các nghiệm thuộc khoảng (-π/2;π/2) của pt 4sin²2x-1=0 A.0 B. π/6 C. π/3 D. π Câu 4: tìm tổng các nghiệm của pt cos(x+π/4)=1/2 trong khoảng (-π;π) A. π/2 B. -π/2 C. -3π/2 D. π/4
Giải các pt sau:
a) \(\cos^2x-\cos x=0\)
b) \(2\sin2x\) + \(\sqrt{2}\sin4x=0\)
c) \(8\cos^2x+2\sin x-7=0\)
d) \(4\cos^4x+\cos^2x-3=0\)
e) \(\sqrt{3}\tan x-6\cot x+\left(2\sqrt{3}-3\right)=0\)
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)
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a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\) (k ∈ Z)
Vậy...
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)
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c, \(8cos^2x+2sinx-7=0\)
\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)
\(\Leftrightarrow8sin^2x-2sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
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d, \(4cos^4x+cos^2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow cos2x=\dfrac{1}{2}\)
\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)
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e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))
\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)
\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)
Vậy...
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c, \(8cos^2x+2sinx-7=0\)
\(\Leftrightarrow-8sin^2x+2sinx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\)
Với \(sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
Với \(sinx=-\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
d, \(4cos^4x+cos^2x-3=0\)
\(\Leftrightarrow\left(4cos^2x-3\right)\left(cos^2x+1\right)=0\)
\(\Leftrightarrow4cos^2x-3=0\left(\text{Vì }cos^2x+1>0\right)\)
\(\Leftrightarrow cos^2x=\dfrac{3}{4}\)
\(\Leftrightarrow cosx=\pm\dfrac{\sqrt{3}}{2}\)
Với \(cosx=\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)
Với \(cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi\)
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Xem thêm câu trả lời
Giải các pt sau:
a) tan^2x - cot^2(x-π/4) =0
b) 3cot^2(45°-3/2x) -1=0
4) 4cos^2x - 2(1+căn 2)cosx + căn 2=0
a/ \(\tan^2x-\cot^2\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-1-\frac{1}{\sin^2\left(x-\frac{\pi}{4}\right)}+1=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\sin x.\cos\frac{\pi}{4}-\cos x.\sin\frac{\pi}{4}\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x}=0\)
\(\Leftrightarrow\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x-\cos^2x=0\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}\cos^2x-\sin x.\cos x-\frac{1}{2}\cos^2x=0\)
\(\Leftrightarrow\cos^2x+\sin x.\cos x-\frac{1}{2}=0\)
Đến đây là dễ r nha bn :3
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Giải các pt sau :
\(tan^2x+cot^2x=1+cos^2\left(3x+\frac{\pi}{4}\right)\)
\(cos\left(\frac{2\pi}{3}sinx-\frac{2\pi}{3}\right)=1\)
cot\(\left[\frac{\pi}{4}\left(cosx-1\right)\right]=-1\)
a. ĐKXĐ: ...
Ta có: \(\left\{{}\begin{matrix}VT=\left(tanx-cotx\right)^2+2\ge2\\VP=1+cos^2\left(3x+\frac{\pi}{4}\right)\le2\end{matrix}\right.\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}tanx-cotx=0\\cos^2\left(3x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}cos2x=0\\sin\left(3x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
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b.
\(\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi\)
\(\Leftrightarrow sinx-1=3k\)
\(\Leftrightarrow sinx=3k+1\)
Do \(-1\le sinx\le1\)
\(\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0\)
\(\Rightarrow k=0\)
\(\Rightarrow sinx=1\)
\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
c.
ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{4}\left(cosx-1\right)=-\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow cosx-1=4k-1\)
\(\Leftrightarrow cosx=4k\)
Mà \(-1\le cosx\le1\Rightarrow-1\le4k\le1\)
\(\Rightarrow-\frac{1}{4}\le k\le\frac{1}{4}\Rightarrow k=0\)
\(\Rightarrow cosx=0\)
\(\Rightarrow x=\frac{\pi}{2}+k\pi\)