Question

giải pt : 

\(sinx^2+cosx.cos3x+sin2x.cos2x=0\)

Answer 1

\(sin^2x+cosx.cos3x+sin2x.cos2x=0\)

\(\Leftrightarrow sin^2x+\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x+\dfrac{1}{2}sin4x=0\)

\(\Leftrightarrow sin^2x+\dfrac{1}{2}-sin^2x+\dfrac{1}{2}sin4x+\dfrac{1}{2}cos4x=0\)

\(\Leftrightarrow sin4x+cos4x=-1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}4x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\4x+\dfrac{\pi}{4}=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)