Question

giải phương trình:

\(2\cos^3x-\sin2x\sin x=-2\sqrt{2}\cos\left(x+\frac{2019\pi}{4}\right)\)

Answer 1

\(2cos^3x-sin2x.sinx=-2\sqrt{2}cos\left(x-\frac{\pi}{4}+505\pi\right)\)

\(\Leftrightarrow cos^3x-sin^2x.cosx=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)

\(\Leftrightarrow cosx\left(cos^2x-sin^2x\right)=sinx+cosx\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(cos^2x-sinx.cosx\right)=sinx+cosx\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-cos^2x+sinx.cosx\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+sinx.cosx\right)=0\)

\(\Leftrightarrow sinx\left(sinx+cosx\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\sinx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=k\pi\end{matrix}\right.\)