Question

cos(2x+\(\dfrac{\pi}{4}\))+cos(2x-\(\dfrac{\pi}{4}\))+4sinx=2+\(\sqrt{2}\)(1-sinx)

Answer 1

\(\Leftrightarrow2cos2x.cos\left(\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}-\sqrt{2}sinx\)

\(\Leftrightarrow\sqrt{2}cos2x+\left(4+\sqrt{2}\right)sinx=2+\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\left(1-2sin^2x\right)+\left(4+\sqrt{2}\right)sinx=2+\sqrt{2}\)

\(\Leftrightarrow-2\sqrt{2}sin^2x+\left(4+\sqrt{2}\right)sinx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\sqrt{2}>1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)