Solution:
Given:
[tex]sin(x+\frac{3\pi}{2})=\frac{\sqrt{3}}{2\text{ }}\text{ on the interval \lbrack0, 2}\pi][/tex][tex]\begin{gathered} (x+\frac{3\pi}{2})=sin^{-1}(\frac{\sqrt{3}}{2}) \\ x+\frac{3\pi}{2}=\frac{\pi}{3}\text{ or }\frac{2\pi}{3} \\ \\ When\text{ x+}\frac{3\pi}{2}=\frac{\pi}{3} \\ x=\text{ }\frac{\pi}{3}-\frac{3\pi}{2} \\ x=\frac{2\pi-9\pi}{6}=\frac{-7\pi}{6} \\ x=\frac{2\pi}{1}\text{ -}\frac{7\pi}{6} \\ \\ x=\frac{12\pi-7\pi}{6} \\ x=\frac{5\pi}{6} \end{gathered}[/tex][tex]\begin{gathered} When\text{ x + }\frac{3\pi}{2}=\frac{2\pi}{3} \\ \\ x=\frac{2\pi}{3}-\frac{3\pi}{2} \\ x=\frac{4\pi-9\pi}{6} \\ x=\frac{-5\pi}{6} \\ x=\frac{2\pi}{1}\text{ -}\frac{5\pi}{6} \\ x=\frac{12\pi-5\pi}{6} \\ x=\frac{7\pi}{6} \end{gathered}[/tex]
Thus, The solution to the equation is
[tex]\lbrace\frac{5\pi}{6},\text{ }\frac{7\pi}{6}\rbrace[/tex]
