Question

Решите уравнение 16x^2-40x+27=(sqrt2-sin(8Пx/5))(sqrt2+sin(8Пx/5))

Answer 1

$<var>16x^2-40x+27=(\sqrt2-\sin(8\pi x/5))(\sqrt2+\sin(8\pi x/5))\\ 16x^2-40x+25+2=2-\sin^2(8\pi x/5)\\ (4x-5)^2=-\sin^2(8\pi x/5)\\ \begin{cases}4x-5=0\\ \sin(8\pi x/5)=0\end{cases}\\x=\frac54</var>$