Question

\(\cos^2x-\sin^2x=\sin3x+\cos4x\)

Answer 1

\(\Leftrightarrow cos2x=sin3x+cos4x\)

\(\Leftrightarrow sin3x+\left(cos4x-cos2x\right)=0\)

\(\Leftrightarrow sin3x-2sin3x.sinx=0\)

\(\Leftrightarrow sin3x\left(1-2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{3}\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Answer 2

\(cos^2x-sin^2x=sin3x+cos4x\\ \Leftrightarrow sin3x+cos4x-cos2x=0\\ \Leftrightarrow sin3x-2sin3x\cdot sinx=0\\ \Leftrightarrow\left[{}\begin{matrix}sin3x=0\\sinx=\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{3}\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}\pi+k2\pi\end{matrix}\right.\)