giải phương trình lượng giác

D

demon311

$\cos 2x-2\sqrt{ 3}\sin x \cos x=2\sin x\\
\leftrightarrow \cos 2x-\sqrt{ 3}\sin 2x=2\sin x \\
\leftrightarrow \dfrac{ 1}{2}\cos 2x-\dfrac{ \sqrt{ 3}}{2}\sin x=\sin x\\
\leftrightarrow \sin (\dfrac{ \pi}{6}-2x)=\sin x \\
\leftrightarrow \left[ \begin{array}{ll}
\dfrac{ \pi}{6}-2x=x +k2\pi \\
\dfrac{ \pi}{6}-2x=\pi-x+k2\pi
\end{array} \right. \\
\leftrightarrow \left[ \begin{array}{ll}
x=\dfrac{ \pi}{18}+k\dfrac{ 2\pi}{3} \\
x=\dfrac{ -5\pi}{6}+k2\pi
\end{array} \right. $
 
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