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)
