\alpha\in(0;\frac{\pi}{2})\\
\alpha=(-1)^n\arcsin\frac{3}{5}+\pi n, n\in Z;\\
n=0;\\
\alpha=\arcsin\frac{3}{5}\in(0;\frac{\pi}{2});\\
n=1;\\
\alpha=\pi-\arcsin\frac{3}{5}\notin(0;\frac{\pi}{2});\\
n=-1;\\
\alpha=-\pi-\arcsin\frac{3}{5}\notin(0;\frac{\pi}{2});\\
\\
\\
" alt="\sin\alpha=\frac{3}{5};\ \ \ 0<\alpha<\frac{\pi}{2}==>\alpha\in(0;\frac{\pi}{2})\\
\alpha=(-1)^n\arcsin\frac{3}{5}+\pi n, n\in Z;\\
n=0;\\
\alpha=\arcsin\frac{3}{5}\in(0;\frac{\pi}{2});\\
n=1;\\
\alpha=\pi-\arcsin\frac{3}{5}\notin(0;\frac{\pi}{2});\\
n=-1;\\
\alpha=-\pi-\arcsin\frac{3}{5}\notin(0;\frac{\pi}{2});\\
\\
\\
" align="absmiddle" class="latex-formula">
значит
Ответ:
