Question

Giair pt: \(sinx+4cosx=2+sin2x\)

Answer 1

\(\Leftrightarrow sinx+4cosx=2+2sinx.cosx\)

\(\Leftrightarrow\left(sinx-2\right)-2cosx\left(sinx-2\right)=0\)

\(\Leftrightarrow\left(sinx-2\right)\left(1-2cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=2\left(vn\right)\\1=2cosx\end{matrix}\right.\)\(\Rightarrow cosx=\dfrac{1}{2}\)

\(\Leftrightarrow cosx=cos\dfrac{\pi}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{-\pi}{3}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

Answer 2