`cos (x - (2π)/3) + sin2x = 0`
`⇔cos (x - (2π)/3)=-sin(2x)`
`⇔cos (x - (2π)/3)=cos (π/2 + 2x)`
`⇔` $\left[\begin{matrix} x-\dfrac{2π}{3}=\dfrac{π}{2}+2x+k2π\\ x-\dfrac{2π}{3}=-\dfrac{π}{2}-2x+k2π\end{matrix}\right.$
`⇔` $\left[\begin{matrix} x=-\dfrac{7π}{6}+k2π\\ x=\dfrac{π}{18}+k\dfrac{2π}{3}\end{matrix}\right.$