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

=>x/3-pi/4=x+k2pi hoặc x/3-pi/4=pi-x+k2pi

=>2/3x=-pi/4+k2pi hoặc 4/3x=5/4pi+k2pi

=>x=-3/8pi+k3pi hoặc x=15/16pi+k*3/2pi

b: =>(sin3x-sin2x)(sin3x+sin2x)=0

=>sin3x-sin2x=0 hoặc sin 3x+sin 2x=0

=>sin 3x=sin 2x hoặc sin 3x=sin(-2x)

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

=>x=k2pi hoặc x=pi/5+k2pi/5 hoặc x=k2pi/5 hoặc x=pi+k2pi

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-7\pi}{12}+k2\pi\\4x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7\pi}{24}+k\pi\\x=\dfrac{13\pi}{48}+k\pi\end{matrix}\right.\left(k\in Z\right)\)

\(sin\left(3x+\dfrac{\Pi}{4}\right)=sin\left(x-\dfrac{\Pi}{3}\right)\)

\(\Leftrightarrow3x+\dfrac{\Pi}{4}=x-\dfrac{\Pi}{3}+K2\Pi\)

\(\Leftrightarrow2x=-\dfrac{7\Pi}{12}+K2\Pi\)      

\(\Leftrightarrow x=-\dfrac{7\Pi}{24}+K\Pi\)           \(\left(K\in Z\right)\)

\(\sqrt{3}cos\left(x+\dfrac{\pi}{2}\right)+sin\left(x-\dfrac{\pi}{2}\right)=2sin2x\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{2}\right)-\dfrac{1}{2}cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{2}+x\right)=sin2x\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx+sin2x=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)+sin2x=0\)

\(\Leftrightarrow2sin\left(\dfrac{3x}{2}+\dfrac{\pi}{12}\right).cos\left(\dfrac{\pi}{12}-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(\dfrac{3x}{2}+\dfrac{\pi}{12}\right)=0\\cos\left(\dfrac{\pi}{12}-\dfrac{x}{2}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}+\dfrac{\pi}{12}=k\pi\\\dfrac{\pi}{12}-\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=-\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

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

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

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

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

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

a: sin x=-6/5=-1,2

mà -1<=sin x<=1

nên \(x\in\varnothing\)
b: sin3x=căn 3/2

=>3x=pi/3+k2pi hoặc 3x=2/3pi+k2pi

=>x=pi/9+k2pi/3 hoặc x=2/9pi+k2pi/3

c: \(sin\left(x+\dfrac{pi}{3}\right)=sin\left(\dfrac{3}{4}pi\right)\)

=>x+pi/3=3/4pi+k2pi hoặc x+pi/3=1/4pi+k2pi

=>x=5/12pi+k2pi hoặc x=-1/12pi+k2pi

d: =>sin(x+5/6pi)=5/4

mà sin(x+5/6pi) thuộc [-1;1]

nên \(x\in\varnothing\)

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

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

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

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

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

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

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

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

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

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

Chỉ II đúng

`a)sin x =4/3`

`=>` Ptr vô nghiệm vì `-1 <= sin x <= 1`

`b)sin 2x=-1/2`

`<=>[(2x=-\pi/6+k2\pi),(2x=[7\pi]/6+k2\pi):}`

`<=>[(x=-\pi/12+k\pi),(x=[7\pi]/12+k\pi):}`    `(k in ZZ)`

`c)sin(x - \pi/7)=sin` `[2\pi]/7`

`<=>[(x-\pi/7=[2\pi]/7+k2\pi),(x-\pi/7=[5\pi]/7+k2\pi):}`

`<=>[(x=[3\pi]/7+k2\pi),(x=[6\pi]/7+k2\pi):}`     `(k in ZZ)`

`d)2sin (x+pi/4)=-\sqrt{3}`

`<=>sin(x+\pi/4)=-\sqrt{3}/2`

`<=>[(x+\pi/4=-\pi/3+k2\pi),(x+\pi/4=[4\pi]/3+k2\pi):}`

`<=>[(x=-[7\pi]/12+k2\pi),(x=[13\pi]/12+k2\pi):}`    `(k in ZZ)`

a: sin x=4/3

mà -1<=sinx<=1

nên \(x\in\varnothing\)

b: sin 2x=-1/2

=>2x=-pi/6+k2pi hoặc 2x=7/6pi+k2pi

=>x=-1/12pi+kpi và x=7/12pi+kpi

c: \(sin\left(x-\dfrac{pi}{7}\right)=sin\left(\dfrac{2}{7}pi\right)\)

=>x-pi/7=2/7pi+k2pi hoặc x-pi/7=6/7pi+k2pi

=>x=3/7pi+k2pi và x=pi+k2pi

d: 2*sin(x+pi/4)=-căn 3

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

=>x+pi/4=-pi/3+k2pi hoặc x-pi/4=4/3pi+k2pi

=>x=-7/12pi+k2pi hoặc x=19/12pi+k2pi

a, \(\left|sinx+\dfrac{1}{2}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow sin^2x+sinx+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow sin^2x+sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)

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

b, \(tan^2\left(x+\dfrac{\pi}{6}\right)=3\)

\(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=\pm\sqrt{3}\)

\(\Leftrightarrow x+\dfrac{\pi}{6}=\pm\dfrac{\pi}{3}+k\pi\)

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

c, \(2sin\left(4x-\dfrac{\pi}{3}\right)-1=0\)

\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\4x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}{2}\end{matrix}\right.\)