Question

1) Giải phương trình \(\cos2x+5=2\left(2-cosx\right)\left(\sin x-cosx\right)\)

Answer 1

\(\Leftrightarrow2cos^2x+4=4\left(sinx-cosx\right)-2sinx.cosx+2cos^2x\)

\(\Leftrightarrow4\left(sinx-cosx\right)-2sinx.cosx+1-5=0\)

\(\Leftrightarrow4\left(sinx-cosx\right)+\left(sinx-cosx\right)^2-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=1\\sinx-cosx=-5\left(voli\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=sin\frac{\pi}{4}\)

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