Sin^2x+2sin x cos x -3 cos^2x = 0

Ответы 1

$sin^2x+2sinx*cosx-3cos^2x=0$ Однородное уравнение 2-го порядка. Делим обе части на cos^2x $tg^2x+2tgx-3=0$ $tgx=t, t eq \frac{\pi}2+\pi n; ne Z$ $t^2+2t-3=0$ $D=4+12=16$ $t_1=\frac{-2+4}2=1$ $t_2=\frac{-2-4}2=-3$ $\left[\begin{array}{ccc}tgx=1\\tgx=-3\end{array}ight = \left[\begin{array}{ccc}x=\frac{\pi}4+\pi n;n\in Z\\x=-arctg(3)+\pi n;n \in Z\end{array}ight$ if (x_square % 5 >= 3) x_square = x_square / 5 + 1; else x_square = x_square / 5; x_square = x_square * 5; if (y_square % 5 >= 3) y_square = y_square / 5 + 1; else y_square = y_square / 5; y_square = y_square * 5; BORDER = false; if ((x_square >= x_diod - 15) && (y_square >= y_diod - 45) && (x_square <= x_diod + 15) && (y_square <= y_diod + 40)) { BORDER_DIOD = false; } else BORDER_DIOD = true; if ((x_square >= x_source - 30) && (y_square >= y_source - 40) && (x_square <= x_source + 11) && (y_square <= y_source + 40)) { BORDER_SOURCE = false; } else BORDER_SOURCE = true; InvalidateRect(hWnd, NULL, false);
- Автор:
  mochag73t
- 6 лет назад
- 0
Добавить свой ответ