`sin(5x+π)=cos(2x-\frac{π}{3})`
⇔ `sin(5x+π)=sin(\frac{π}{2}-2x+\frac{π}{3})`
⇔ `sin(5x+π)=sin(\frac{5π}{6}-2x)`
⇔ $\left [\begin{array}{l} 5x+π=\dfrac{5π}{6}-2x+k2π \\ 5x+π=\pi-\dfrac{5π}{6}+2x+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} 7x=-\dfrac{\pi}{6}+k2π \\ 3x=-\dfrac{5π}{6}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=-\dfrac{\pi}{42}+\dfrac{k2π}{7} \\ x=-\dfrac{5π}{18}+\dfrac{k2π}{3} \end{array} \right. \ (k∈\mathbb{Z})$
`T=-\frac{\pi}{42}+\frac{k2π}{7}-\frac{5π}{18}+\frac{k2π}{3}=-\frac{19π}{63}+\frac{k20π}{21}`
