2sin x*cos x - 2√3*(cos x*cos(7pi/6) + sin x*sin(7pi/6)) = 3cos x2sin x*cos x - 2√3*(cos x*(-√3/2) + sin x*(-1/2)) = 3cos x2sin x*cos x + 2√3*√3/2*cos x + 2√3/2*sin x = 3cos x2√3*√3/2*cos x = 3cos x, их можно сократить.2sin x*cos x + √3*sin x = 0sin x*(2cos x + √3) = 01) sin x = 0; x = pi*k. На отрезке [-3pi/2; 0] будут корни x1 = -pi; x2 = 02) cos x = -√3/2; x = +-5pi/6 + 2pi*n. На отрезке [-3pi/2; 0] будет x3 = -5pi/6Ответ: x1 = -pi; x2 = 0; x3 = -5pi/6
#include <cstdio>
using namespace std;
class Clock
{
int h, m, s;
int isPM(int hour) {
if ( (hour > 12) && (hour < 24) )
return 1;
else
return 0;
}
public:
Clock() : h(0), m(0), s(0) {}
void setTime(int ph, int pm, int ps) {
try {
if ( !( (ph < 0) || (ph > 23) ) )
h = ph;
else
throw "Неверный час";
if ( !( (pm < 0) || (pm > 59) ) )
m = pm;
else
throw "Неверные минуты";
if ( !( (ps < 0) || (ps > 59) ) )
s = ps;
else
throw "Неверные секунды";
} catch (const char* e) {
printf("Ошибка: %s", e);
}
}
Clock(int ph, int pm, int ps) {
setTime(ph, pm, ps);
}
int hour() { return h; }
int min() { return m; }
int sec() { return s; }
void print1() {
printf("%d часов %d минут %d секунд\n", h, m, s);
}
void print2() {
char a_p;
int ph;
if (isPM(h)) {
a_p = 'p';
ph = h - 12;
} else {
a_p = 'a';
ph = h;
}
printf("%d %c. m. %d минут %d секунд\n", ph, a_p, m, s);
}
};
/* Проверка некоторых функций */
int main()
{
Clock clock1(23, 11, 07), clock2;
clock1.print2();
clock2.print1();
return 0;
}
Ответ: 32
Решение:
var arr: array [1..12] of real; s, tmp: real; element_id_1, element_id_2: integer;begin s:=0; for element_id_1:=1 to 12 do readln (arr[element_id_1]); for element_id_1:=1 to 12 do begin for element_id_2:=1 to element_id_1-1 do begin if arr[element_id_2]>arr[element_id_1] then begin tmp:=arr[element_id_2]; arr[element_id_2]:=arr[element_id_1]; arr[element_id_1]:=tmp; s:=s+1; end; end; end; writeln(s);end.
Var
x0, xk, h, e, x, a, s: real;
i: integer;
begin
readln(x0, xk, h, e);
x := x0;
repeat
a := 2 * x;
s := a;
i := 0;
repeat
i := i + 1;
a := a * 4 * x * x / (2 * i - 1) / (2 * i);
s := s + a;
until abs(a) < e;
writeln(s:15:5, ' ', Sinh(2*x):15:5);
x := x + h;
until x >= xk;
end.