Question

Giải các phương trình sau :

a) \(\sin\left(x-\frac{2\pi}{3}\right)=\cos2x\)

b) \(\frac{\sin^3x+\cos^3x}{2\cos x-\sin x}=\cos2x\)

Answer 1

a) Ta có : \(sin\left(x-\frac{2\pi}{3}\right)=cos2x\)

\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2\pi}{3}=\frac{\pi}{2}-2x+k2\pi\\x-\frac{2\pi}{3}=\pi-\frac{\pi}{2}+2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{18}+k\frac{2\pi}{3}\\x=-\frac{7\pi}{6}-k2\pi\end{matrix}\right.\)

Vậy ...