Question

Giải pt sau:

\(sin4x+1=2sinx\left(1+cosx\right)+cos3x\)

Answer 1

\(sin4x+1-2sinx-sin2x-cos3x=0\)

\(\Leftrightarrow2cos3x.sinx-cos3x+1-2sinx=0\)

\(\Leftrightarrow cos3x\left(2sinx-1\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(cos3x-1\right)\left(2sinx-1\right)=0\Rightarrow\left[{}\begin{matrix}cos3x=1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)