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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(sinx=sin\left(\dfrac{\Omega}{4}\right)\)
=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{4}+k2\Omega\\x=\Omega-\dfrac{\Omega}{4}+k2\Omega=\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
b: cos2x=cosx
=>\(\left[{}\begin{matrix}2x=x+k2\Omega\\2x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\3x=k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{k2\Omega}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{k2\Omega}{3}\)
c:
ĐKXĐ: \(x-\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{5}{6}\Omega+k\Omega\)
\(tan\left(x-\dfrac{\Omega}{3}\right)=\sqrt{3}\)
=>\(x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)
=>\(x=\dfrac{2}{3}\Omega+k\Omega\)
d:
ĐKXĐ: \(2x+\dfrac{\Omega}{6}< >k\Omega\)
=>\(2x< >-\dfrac{\Omega}{6}+k\Omega\)
=>\(x< >-\dfrac{1}{12}\Omega+\dfrac{k\Omega}{2}\)
\(cot\left(2x+\dfrac{\Omega}{6}\right)=cot\left(\dfrac{\Omega}{4}\right)\)
=>\(2x+\dfrac{\Omega}{6}=\dfrac{\Omega}{4}+k\Omega\)
=>\(2x=\dfrac{1}{12}\Omega+k\Omega\)
=>\(x=\dfrac{1}{24}\Omega+\dfrac{k\Omega}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: 2*sin x+1<>0
=>sin x<>-1/2
=>x<>-pi/6+k2pi và x<>7/6pi+k2pi
b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)
mà 1+cosx>=0
nên 2-cosx>=0
=>cosx<=2(luôn đúng)
c ĐKXĐ: tan x>0
=>kpi<x<pi/2+kpi
d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)
=>cos(x-pi/4)<>1/2
=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi
=>x<>7/12pi+k2pi và x<>-pi/12+k2pi
e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi
=>x<>5/6pi+kpi và x<>kpi-pi/4
f: ĐKXĐ: cos^2x-sin^2x<>0
=>cos2x<>0
=>2x<>pi/2+kpi
=>x<>pi/4+kpi/2
![](https://rs.olm.vn/images/avt/0.png?1311)
d/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+\frac{5}{2}=4\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow2sin^2\left(x+\frac{\pi}{6}\right)+4sin\left(x+\frac{\pi}{6}\right)-\frac{7}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{-2+\sqrt{11}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{-2-\sqrt{11}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\\x=\frac{5\pi}{6}-arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)