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

angelochekbobi angelochekbobi    3   04.09.2019 05:40    1

Ответы
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
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