Question

4(sin⁴x+cos⁴x) +√3 sin(4x+5π)=2

Answer 1

\(4\left(sin^2x+cos^2x\right)^2-8sin^2xcos^2x-\sqrt{3}sin4x=2\)

\(\Leftrightarrow2-2sin^22x-\sqrt{3}sin4x=0\)

\(\Leftrightarrow1+cos4x-\sqrt{3}sin4x=0\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=\frac{1}{2}\)

\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\4x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)