Question

cos(x+pi/3)+sin2x=0

Answer 1

\(\Leftrightarrow sin2x=-cos\left(x+\dfrac{pi}{3}\right)\\ \Leftrightarrow sin2x=sin\left(x+\dfrac{pi}{3}-\dfrac{pi}{2}\right)\\ \Leftrightarrow sin2x=sin\left(x+\dfrac{5pi}{6}\right)\\ \Leftrightarrow|\left[{}\begin{matrix}2x=x+\dfrac{5pi}{6}+k2pi\\2x=pi-x-\dfrac{5pi}{6}+k2pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5pi}{6}+k2pi\\3x=pi-\dfrac{5pi}{6}+k2pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5pi}{6}+k2pi\\x=\dfrac{pi}{18}-\dfrac{k2pi}{3}\end{matrix}\right.\)