Question

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

Answer 1

$sin^2x+2sinx*cosx-3cos^2x=0$

Однородное уравнение 2-го порядка. Делим обе части на cos^2x

$tg^2x+2tgx-3=0$
$tgx=t, t neq frac{pi}2+pi n; ne Z$
$t^2+2t-3=0$
$D=4+12=16$
$t_1=frac{-2+4}2=1$
$t_2=frac{-2-4}2=-3$

$left[begin{array}{ccc}tgx=1\tgx=-3end{array}right = left[begin{array}{ccc}x=frac{pi}4+pi n;nin Z\x=-arctg(3)+pi n;n in Zend{array}right$


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);