\(cos\left(2x+\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow2cos\dfrac{3x}{2}.cos\left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{3x}{2}=0\\cos\left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\\\dfrac{x}{2}+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

a: \(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)+\sqrt{3}=0\)

=>\(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)=-\sqrt{3}\)

=>\(sin\left(x+\dfrac{\Omega}{5}\right)=-\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{5}=-\dfrac{\Omega}{3}+k2\Omega\\x+\dfrac{\Omega}{5}=\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{8}{15}\Omega+k2\Omega\\x=\dfrac{4}{3}\Omega-\dfrac{\Omega}{5}+k2\Omega=\dfrac{17}{15}\Omega+k2\Omega\end{matrix}\right.\)

b: \(sin\left(2x-50^0\right)=\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}2x-50^0=60^0+k\cdot360^0\\2x-50^0=300^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=110^0+k\cdot360^0\\2x=350^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=55^0+k\cdot180^0\\x=175^0+k\cdot180^0\end{matrix}\right.\)

c: \(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)-1=0\)

=>\(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)=1\)

=>\(tan\left(2x-\dfrac{\Omega}{3}\right)=\dfrac{1}{\sqrt{3}}\)

=>\(2x-\dfrac{\Omega}{3}=\dfrac{\Omega}{6}+k2\Omega\)

=>\(2x=\dfrac{1}{2}\Omega+k2\Omega\)

=>\(x=\dfrac{1}{4}\Omega+k\Omega\)

1.

\(\Leftrightarrow cos\left(2x+\dfrac{4\pi}{3}\right)=0\)

\(\Leftrightarrow2x+\dfrac{4\pi}{3}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow2x=-\dfrac{5\pi}{6}+k\pi\)

\(\Leftrightarrow x=-\dfrac{5\pi}{12}+\dfrac{k\pi}{2}\)

b.

\(\Leftrightarrow2+2cos\left(2x+\dfrac{\pi}{3}\right)-3=0\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\2x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

c.

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{6}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=k\pi\end{matrix}\right.\)

1: cos(2x+pi/6)=cos(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=3x-pi/3+k2pi

=>5x=pi/6+k2pi hoặc -x=-1/2pi+k2pi

=>x=pi/30+k2pi/5 hoặc x=pi-k2pi

2: sin(2x+pi/6)=sin(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=pi-pi/3+3x+k2pi

=>5x=pi/6+k2pi hoặc -x=2/3pi-pi/6+k2pi

=>x=pi/30+k2pi/5 hoặc x=-1/2pi-k2pi

1) \(cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{3}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-3x+k2\pi\\2x+\dfrac{\pi}{6}=-\dfrac{\pi}{3}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{3}-\dfrac{\pi}{6}+k2\pi\\3x-2x=\dfrac{\pi}{3}+\dfrac{\pi}{6}-k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\) \(\left(k\in N\right)\)

d: cos^2x=1

=>sin^2x=0

=>sin x=0

=>x=kpi

a: =>sin 4x=cos(x+pi/6)

=>sin 4x=sin(pi/2-x-pi/6)

=>sin 4x=sin(pi/3-x)

=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi

=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3

b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi

=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi

c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi

=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2

\(cos\left(x-\dfrac{\pi}{3}\right)=sin\left(2x+\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{\pi}{3}-x+l2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{9}+l\dfrac{2\pi}{3}\end{matrix}\right.\)

Chỉ II đúng

Pt \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}-x=2x+\dfrac{\pi}{3}+k2\pi\\\dfrac{\pi}{3}-x=-2x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-k2\pi}{3}\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...

Pt \(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{4}=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\),\(k\in Z\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{19\pi}{24}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...
Hôm qua họ bảo toi ra lấy CCCD nma toi chưa đi, nay toi đi họ lại đang họp, liệu mai toi đi có bị ăn chửi ko, mn cho ý kiến đi :<

\(2sin\left(2x-\dfrac{\pi}{4}\right)+\sqrt{3}=0\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=sin\left(-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{12}+k2\pi\\2x=\dfrac{19\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{19\pi}{24}+k\pi\end{matrix}\right.\)