Question

\(sin^2x+sinx.cos4x+cos^24x=\frac{3}{4}\)

Answer 1

\(\Leftrightarrow sin^2x+sinx.cos4x+\frac{1}{4}cos^24x=\frac{3}{4}\left(1-cos^24x\right)\)

\(\Leftrightarrow\left(sinx+\frac{1}{2}cos4x\right)^2=\frac{3}{4}sin^24x\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\frac{1}{2}cos4x=\frac{\sqrt{3}}{2}sin4x\\sinx+\frac{1}{2}cos4x=-\frac{\sqrt{3}}{2}sin4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x\\-sinx=\frac{\sqrt{3}}{2}sin4x+\frac{1}{2}cos4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=sin\left(4x-\frac{\pi}{6}\right)\\sin\left(-x\right)=sin\left(4x+\frac{\pi}{6}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\) (dạng cơ bản \(sinx=sina\) bạn tự giải nốt)