\(3\cot\left(2x-80^0\right)-\sqrt{3}=0\)
Giải PT:
a1. \(\cot\left(2x+\dfrac{\pi}{3}\right)\)=\(-\sqrt{3}\)
a2. \(\cot\left(3x-10^{\cdot}\right)\cot2x=1\)
a3. \(\cot\left(\dfrac{\pi}{4}-2x\right)-\tan x=0\)
a4. \(\cot\left(30^{\cdot}+3x\right)+\tan\left(x-10^{\cdot}\right)=0\)
a1.
$\cot (2x+\frac{\pi}{3})=-\sqrt{3}=\cot \frac{-\pi}{6}$
$\Rightarrow 2x+\frac{\pi}{3}=\frac{-\pi}{6}+k\pi$ với $k$ nguyên
$\Leftrightarrow x=\frac{-\pi}{4}+\frac{k}{2}\pi$ với $k$ nguyên
a2. ĐKXĐ:...............
$\cot (3x-10^0)=\frac{1}{\cot 2x}=\tan 2x$
$\Leftrightarrow \cot (3x-\frac{\pi}{18})=\cot (\frac{\pi}{2}-2x)$
$\Rightarrow 3x-\frac{\pi}{18}=\frac{\pi}{2}-2x+k\pi$ với $k$ nguyên
$\Leftrightarrow x=\frac{\pi}{9}+\frac{k}{5}\pi$ với $k$ nguyên.
a3. ĐKXĐ:........
$\cot (\frac{\pi}{4}-2x)-\tan x=0$
$\Leftrightarrow \cot (\frac{\pi}{4}-2x)=\tan x=\cot (\frac{\pi}{2}-x)$
$\Rightarrow \frac{\pi}{4}-2x=\frac{\pi}{2}-x+k\pi$ với $k$ nguyên
$\Leftrightarrow x=-\frac{\pi}{4}+k\pi$ với $k$ nguyên.
a4. ĐKXĐ:.....
$\cot (\frac{\pi}{6}+3x)+\tan (x-\frac{\pi}{18})=0$
$\Leftrightarrow \cot (\frac{\pi}{6}+3x)=-\tan (x-\frac{\pi}{18})=\tan (\frac{\pi}{18}-x)$
$=\cot (x+\frac{4\pi}{9})$
$\Rightarrow \frac{\pi}{6}+3x=x+\frac{4\pi}{9}+k\pi$ với $k$ nguyên
$\Rightarrow x=\frac{5}{36}\pi + \frac{k}{2}\pi$ với $k$ nguyên.
1) nghiệm dương nhỏ nhất của phương trình \(cot\left(x-\dfrac{\pi}{6}\right)=\sqrt{3}\) là
2) phương trình \(sin\left(\dfrac{2x}{3}+\dfrac{\pi}{3}\right)=0\) có nghiệm là
3) họ nghiệm của phương trình \(cot\)(2x - 30 độ) = \(\sqrt{3}\) là
Giải các phương trình sau:
a) \(\sin x = \frac{{\sqrt 3 }}{2}\);
b) \(2\cos x = - \sqrt 2 \);
c) \(\sqrt 3 \tan \left( {\frac{x}{2} + {{15}^0}} \right) = 1\);
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\)
a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)
b) \(2\cos x = - \sqrt 2 \;\; \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x = - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)
c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)
\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
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, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)
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, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)
Vậy...
c, \(8cos^2x+2sinx-7=0\)
\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)
\(\Leftrightarrow8sin^2x-2sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
Vậy...
d, \(4cos^4x+cos^2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow cos2x=\dfrac{1}{2}\)
\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)
Vậy...
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...
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 \(cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi\)
Giải các phương trình :
a) \(\tan\left(2x+45^0\right)=-1\)
b) \(\cot\left(x+\dfrac{\pi}{3}\right)=\sqrt{3}\)
c) \(\tan\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=\tan\dfrac{\pi}{8}\)
d) \(\cot\left(\dfrac{x}{3}+20^0\right)=-\dfrac{\sqrt{3}}{3}\)
a) \(x=-45^0+k90^0,k\in\mathbb{Z}\)
b) \(x=-\dfrac{\pi}{6}+k\pi,k\in\mathbb{Z}\)
c) \(x=\dfrac{3\pi}{4}+k2\pi,k\in\mathbb{Z}\)
d) \(x=300^0+k540^0,k\in\mathbb{Z}\)
giải phương trình:
a,\(\left(cot\frac{x}{3}-1\right)\left(cot\frac{x}{2}+1\right)=0\)
b,\(tan\left(x-30\text{° }\right)cos\left(2x-150\text{° }\right)=0\)
c,\(\left(3tanx+\sqrt{3}\right)\left(2sinx-1\right)=0\)
d, \(cos2x.cot\left(x-\frac{\pi}{4}\right)=0\)
Giải PT
a) \(\left|\sin x-\cos x\right|+\left|\sin x+\cos x\right|=2\)
b) \(\tan x-3\cot x=4\left(\sin x+\sqrt{3}\cos x\right)\)
c) \(2\sin^2x-2\sqrt{2}\sin x+3\tan^22x-2\sqrt{3}\tan2x+2=0\)
a) \(\left|sinx-cosx\right|+\left|sinx+cosx\right|=2\)
\(\Leftrightarrow\left(sinx-cosx\right)^2+2\left|sinx-cosx\right|\left|sinx+cosx\right|+\left(cosx+sinx\right)^2=4\)
\(\Leftrightarrow2\left(sin^2x+cos^2x\right)+2\left|\left(sinx-cosx\right)\left(sinx+cosx\right)\right|=4\)
\(\Leftrightarrow\left|sin^2x-cos^2x\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=1\\sin^2x-cos^2x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=sin^2x+cos^2x\\sin^2x-cos^2x=-\left(sin^2x+cos^2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=0\\sin^2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=0\end{matrix}\right.\)\(\Rightarrow cosx.sinx=0\Rightarrow sin2x=0\)
\(\Rightarrow x=\dfrac{k\pi}{2},k\in Z\)
Vậy...
b) ĐK:\(x\ne\dfrac{k\pi}{2};k\in Z\)
Pt \(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cosx}{sinx}=4\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{cosx.sinx}=4\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\dfrac{\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx\right)}{sinx.cosx}=4\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}cosx=0\left(1\right)\\\dfrac{sinx-\sqrt{3}cosx}{sinx.cosx}=4\left(2\right)\end{matrix}\right.\)
Từ \(\left(1\right)\Leftrightarrow tanx=-\sqrt{3}\Leftrightarrow x=-\dfrac{\pi}{3}+k\pi,k\in Z\)
Từ (2)\(\Leftrightarrow sinx-\sqrt{3}cosx=4sinx.cosx\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=2sinx.cosx\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin2x\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)
c) ĐK: \(x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\left(k\in Z\right)\)
Pt \(\Leftrightarrow\left(\sqrt{2}sinx-1\right)^2+\left(\sqrt{3}tan2x-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}sinx-1=0\\\sqrt{3}tan2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}sinx=\dfrac{1}{\sqrt{2}}\\tan2x=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)
Vậy pt vô nghiệm
giải các pt
a) \(\frac{tanx-8\sqrt{3}}{tanx-2\sqrt{3}}=3\)
b) \(tan2x+cot\frac{5\pi}{8}=0\)
c) \(\sqrt{3}tan\left(\frac{\pi}{9}-2x\right)+3=0\)
d) \(\left(tanx-5\right)\left(tan4-tan2x\right)=0\)
a/
ĐKXĐ: ...
\(\Leftrightarrow tanx-8\sqrt{3}=3tanx-6\sqrt{3}\)
\(\Leftrightarrow2tanx=-2\sqrt{3}\)
\(\Rightarrow tanx=-\sqrt{3}\Rightarrow x=-\frac{\pi}{3}+k\pi\)
b/
\(\Leftrightarrow tan2x=-cot\left(\frac{5\pi}{8}\right)\)
\(\Leftrightarrow tan2x=tan\left(\frac{\pi}{2}+\frac{5\pi}{8}\right)\)
\(\Leftrightarrow tan2x=tan\left(\frac{9\pi}{8}\right)\)
\(\Rightarrow2x=\frac{9\pi}{8}+k\pi\Rightarrow x=\frac{9\pi}{16}+\frac{k\pi}{2}\)
c/
\(\Leftrightarrow\sqrt{3}tan\left(\frac{\pi}{9}-2x\right)=-3\)
\(\Leftrightarrow tan\left(\frac{\pi}{9}-2x\right)=-\sqrt{3}\)
\(\Rightarrow\frac{\pi}{9}-2x=-\frac{\pi}{3}+k\pi\)
\(\Rightarrow x=\frac{2\pi}{9}+\frac{k\pi}{2}\)
d/
\(\Leftrightarrow\left[{}\begin{matrix}tanx=5\\tan2x=tan4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\2x=4+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\x=2+\frac{k\pi}{2}\end{matrix}\right.\)
Giải các phương trình :
a) \(\cos\left(22^0-t\right)\cos\left(82^0-t\right)+\cos\left(112^0-t\right)\cos\left(172^0-t\right)=\dfrac{1}{2}\left(\sin t+\cos t\right)\)
b) \(\sin^2\left(t+45^0\right)-\sin^2\left(t-30^0\right)-\sin15^0\cos\left(2t+15^0\right)=\dfrac{1}{2}\sin6t\)
c) \(\sin^82x+\cos^82x=\dfrac{41}{128}\)
d) \(\sqrt{4\cos^2+1}+\sqrt{4\sin^2x+3}=4\)
e) \(\tan\left(\pi\cot t\right)=\cot\left(\pi\sin t\right)\)