2sin^2(x+3pi/2)=1/2 ПОМОГИТЕ найти решение неравенства

Question 1

Question 2

Это уравнение, а не неравенство, если чё

$\mathtt{2sin^2(x+\frac{3\pi}{2})=\frac{1}{2};~(2sin(x+\frac{3\pi}{2})-1)(2sin(x+\frac{3\pi}{2})+1)=0}$

далее — совокупность: $\mathtt{\left[\begin{array}{ccc}\mathtt{2sin(x+\frac{3\pi}{2})-1=0}\\\mathtt{2sin(x+\frac{3\pi}{2})+1=0}\end{array}\right\to\left[\begin{array}{ccc}\mathtt{sin(x+\frac{3\pi}{2})=\frac{1}{2}}\\\mathtt{sin(x+\frac{3\pi}{2})=-\frac{1}{2}}\end{array}\right}$

решение: $\mathtt{\left[\begin{array}{ccc}\mathtt{\left[\begin{array}{ccc}\mathtt{x+\frac{3\pi}{2}=\frac{\pi}{6}+2n\pi}\\\mathtt{x+\frac{3\pi}{2}=\frac{5\pi}{6}+2n\pi}\end{array}\right}\\\mathtt{\left[\begin{array}{ccc}\mathtt{x+\frac{3\pi}{2}=\frac{7\pi}{6}+2n\pi}\\\mathtt{x+\frac{3\pi}{2}=\frac{11\pi}{6}+2n\pi}\end{array}\right}\end{array}\right\to}$ $\mathtt{\left[\begin{array}{ccc}\mathtt{\left[\begin{array}{ccc}\mathtt{x=2n\pi+\frac{\pi}{6}-\frac{3\pi}{2}=2n\pi-\frac{4\pi}{3}}\\\mathtt{x=2n\pi+\frac{5\pi}{6}-\frac{3\pi}{2}=2n\pi-\frac{2\pi}{3}}\end{array}\right}\\\mathtt{\left[\begin{array}{ccc}\mathtt{x=2n\pi+\frac{7\pi}{6}-\frac{3\pi}{2}=2n\pi-\frac{\pi}{3}}\\\mathtt{x=2n\pi+\frac{11\pi}{6}-\frac{3\pi}{2}=2n\pi+\frac{\pi}{3}}\end{array}\right}\end{array}\right}$

ответ: $x=2n\pi-\frac{4\pi}{3};~2n\pi-\frac{2\pi}{3};~2n\pi-\frac{\pi}{3};~2n\pi+\frac{\pi}{3},~n\in Z$