Question

A)Sin8x=-1/2

 

B)Sin9x+cos9x=0

 

C)Cos5x=√3/2

 

D)Cos5x=sin2x

 

E)Tan(2x+π/3)=√3

 

F)2sin4x+√3=0

 

G)Cos=1/3

 

H)Sin (2x+π/4)=sin(x-π/3)

 

J) cot(3x-2π/3)=-1/√3

Answer 1

\(e,tan\left(2x+\dfrac{\Pi}{3}\right)=\sqrt{3}\Rightarrow tan\left(2x+\dfrac{\Pi}{3}\right)=tan\dfrac{\Pi}{3}\Rightarrow2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{3}+k\Pi\Rightarrow2x=k\Pi\Rightarrow x=\dfrac{k\Pi}{2}\)\(f,2sin4x+\sqrt{3}=0\Rightarrow sin4x=-\dfrac{\sqrt{3}}{2}\Rightarrow sin4x=sin\left(\dfrac{-\Pi}{3}\right)\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{\Pi}{3}+k2\Pi\\4x=\Pi-\left(-\dfrac{\Pi}{3}\right)+k2\Pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-\Pi}{12}+\dfrac{k\Pi}{2}\\x=\dfrac{\Pi}{3}+\dfrac{k\Pi}{2}\end{matrix}\right.\)
\(h,sin\left(2x+\dfrac{\Pi}{4}\right)=sin\left(x-\dfrac{\Pi}{3}\right)\Rightarrow\left[{}\begin{matrix}2x+\dfrac{\Pi}{4}=x-\dfrac{\Pi}{3}+k2\Pi\\2x+\dfrac{\Pi}{4}=\Pi-x+\dfrac{\Pi}{3}+k2\Pi\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7\Pi}{12}+k2\Pi\\3x=\dfrac{13\Pi}{12}+k2\Pi\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7\Pi}{12}+k2\Pi\\x=\dfrac{13\Pi}{36}+\dfrac{k2\Pi}{3}\end{matrix}\right.\)
\(j,cot\left(3x-\dfrac{2\Pi}{3}\right)=\dfrac{-1}{\sqrt{3}}\Rightarrow cot\left(3x-\dfrac{2\Pi}{3}\right)=cot\left(-\dfrac{\Pi}{3}\right)\Rightarrow3x-\dfrac{2\Pi}{3}=\dfrac{-\Pi}{3}+k\Pi\Rightarrow3x=\dfrac{\Pi}{3}+k\Pi\)\(\Rightarrow x=\dfrac{\Pi}{9}+\dfrac{k\Pi}{3}\)

Answer 2

 

a: \(sin8x=-\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}8x=-\dfrac{\Omega}{6}+k2\Omega\\8x=\Omega+\dfrac{\Omega}{6}+k2\Omega=\dfrac{7}{6}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{\Omega}{48}+\dfrac{k\Omega}{4}\\x=\dfrac{7}{48}\Omega+\dfrac{k\Omega}{4}\end{matrix}\right.\)

b: \(sin9x+cos9x=0\)

=>\(\sqrt{2}\cdot sin\left(9x+\dfrac{\Omega}{4}\right)=0\)

=>\(sin\left(9x+\dfrac{\Omega}{4}\right)=0\)

=>\(9x+\dfrac{\Omega}{4}=k\Omega\)

=>\(9x=-\dfrac{\Omega}{4}+k\Omega\)

=>\(x=-\dfrac{\Omega}{36}+\dfrac{k\Omega}{9}\)

c: \(cos5x=\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}5x=\dfrac{\Omega}{6}+k2\Omega\\5x=-\dfrac{\Omega}{6}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{30}+\dfrac{k2\Omega}{5}\\x=-\dfrac{\Omega}{30}+\dfrac{k2\Omega}{5}\end{matrix}\right.\)

d: cos5x=sin2x

=>\(sin2x=sin\left(\dfrac{\Omega}{2}-5x\right)\)

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

=>\(\left[{}\begin{matrix}7x=\dfrac{\Omega}{2}+k2\Omega\\-3x=\dfrac{\Omega}{2}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{14}+\dfrac{k2\Omega}{7}\\x=-\dfrac{\Omega}{6}-\dfrac{k2\Omega}{3}\end{matrix}\right.\)

e:

ĐKXĐ: \(2x+\dfrac{\Omega}{3}\ne\dfrac{\Omega}{2}+k\Omega\)

=>\(2x\ne\dfrac{\Omega}{6}+k\Omega\)

=>\(x\ne\dfrac{\Omega}{12}+\dfrac{k\Omega}{2}\)

 \(tan\left(2x+\dfrac{\Omega}{3}\right)=\sqrt{3}\)

=>\(2x+\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)

=>\(2x=k\Omega\)

=>\(x=\dfrac{k\Omega}{2}\)(nhận)

f: \(2\cdot sin4x+\sqrt{3}=0\)

=>\(2\cdot sin4x=-\sqrt{3}\)

=>\(sin4x=-\dfrac{\sqrt{3}}{2}\)

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

=>\(\left[{}\begin{matrix}x=-\dfrac{\Omega}{12}+\dfrac{k\Omega}{2}\\x=\dfrac{1}{3}\Omega+\dfrac{k\Omega}{2}\end{matrix}\right.\)

g: Sửa đề: \(cosx=\dfrac{1}{3}\)

=>\(\left[{}\begin{matrix}x=arccos\left(\dfrac{1}{3}\right)+k2\Omega\\x=-arccos\left(\dfrac{1}{3}\right)+k2\Omega\end{matrix}\right.\)

h: \(sin\left(2x+\dfrac{\Omega}{4}\right)=sin\left(x-\dfrac{\Omega}{3}\right)\)

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

=>\(\left[{}\begin{matrix}2x-x=-\dfrac{\Omega}{3}-\dfrac{\Omega}{4}+k2\Omega\\2x+x=\dfrac{4}{3}\Omega-\dfrac{1}{4}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{7}{12}\Omega+k2\Omega\\3x=\dfrac{13}{12}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{12}\Omega+k2\Omega\\x=\dfrac{13}{36}\Omega+\dfrac{k2\Omega}{3}\end{matrix}\right.\)

j:

ĐKXĐ: \(3x-\dfrac{2\Omega}{3}\ne k\Omega\)

=>\(3x\ne\dfrac{2}{3}\Omega+k\Omega\)

=>\(x\ne\dfrac{2}{9}\Omega+\dfrac{k\Omega}{3}\)

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

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

=>\(3x=\dfrac{1}{3}\Omega+k\Omega\)

=>\(x=\dfrac{1}{9}\Omega+\dfrac{k\Omega}{3}\)(nhận)