Question

  Giải phương trình lượng giác 

Answer 1

\(sinx-cos\left(x+\dfrac{\Omega}{6}\right)=0\)

=>\(sinx=cos\left(x+\dfrac{\Omega}{6}\right)\)

=>\(sinx=sin\left(\dfrac{\Omega}{2}-x-\dfrac{\Omega}{6}\right)=sin\left(\dfrac{\Omega}{3}-x\right)\)

=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{3}-x+k2\Omega\\x=\Omega-\dfrac{\Omega}{3}+x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Omega}{3}+k2\Omega\\0=\dfrac{2}{3}\Omega+k2\Omega\left(loại\right)\end{matrix}\right.\)

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

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