Question

\(1+cos2x+sinx=2cos^2\frac{x}{2}\)

Answer 1

\(\leftrightarrow2cos^2x+sinx=cosx+1\)

\(\leftrightarrow cos^2x-cosx-sin^2x+sinx=0\)

\(\leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)-\left(cosx-sinx\right)=0\)

\(\leftrightarrow\left(cosx-sinx\right).\left(cosx+sinx-1\right)=0\)

\(cosx-sinx=0\leftrightarrow x=\frac{\pi}{4}+k\pi\left(k\epsilon Z\right)\)

\(cosx+sinx=1\leftrightarrow Sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\leftrightarrow x=k2\pi hoacx=\frac{\pi}{2}+k2\pi\)