Question

giải phương trình √3 cosx - sinx = √2

Answer 1

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{4}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{5\pi}{12}+k2\pi\end{matrix}\right.\)