Question

2 sin7x.sinx + cos8x + \(\sqrt{3}\)sin6x = 1.

Answer 1

\(\Leftrightarrow cos6x-cos8x+cos8x+\sqrt{3}sin6x=1\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x=\frac{1}{2}\)

\(\Leftrightarrow sin\left(6x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k\pi}{3}\\x=\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)