Задача 56899 Помогите пожалуйста...
Условие
Решение
[m]4sin^2x=1[/m]
[m]sin^2x=\frac{1}{4}[/m]
sinx= ± [m]\frac{1}{2}[/m]
1.sinx = [m]\frac{1}{2}[/m]
x_(1) = [m]\frac{π}{6}+2πk[/m]
x_(2) = [m]\frac{5π}{6}+2πk[/m]
2.sinx=[m]-\frac{1}{2}[/m]
x_(1) = [m]\frac{7π}{6}+2πk[/m]
x_(2)=[m]\frac{11π}{6}+2πk[/m]
найдем общие корни:
x_(1) = [m]\frac{π}{6}+πk[/m]
x_(2) = [m]\frac{5π}{6}+πk[/m]
2)[m]4sin^2x-4sinx+1=0[/m]
сделаем замену переменной: sinx = t
4t^(2) -4t+1 = 0
D = (-4)^(2)-4*4*1 = 16-16 = 0
t = [m]\frac{4+0}{8} = \frac{4}{8} = \frac{1}{2}[/m] => sinx = [m]\frac{1}{2}[/m]
x_(1) = [m]\frac{π}{6}+2πk[/m]
x_(2) = [m]\frac{5π}{6}+2πk[/m]
3)[m]2sin^2x-5cosx+1=0[/m]
[m]2(1-cos^2x)-5cosx+1=0[/m]
[m]2-2cos^2x-5cosx+1=0[/m]
[m]-2cos^2x-5cosx+3=0[/m]
[m]2cos^2x+5cosx-3=0[/m]
сделаем замену переменной: cosx = t
2t^(2)+5t-3=0
D = 5^(2)-4*2*(-3) = 49
t_(1) = [m]\frac{-5+7}{4}=\frac{2}{4} = \frac{1}{2}[/m]
t_(2) = [m]\frac{-5-7}{4}=-\frac{12}{4} = -3[/m] - [red]не подходит[/red]
следовательно:
cosx = [m]\frac{1}{2}[/m]
x_(1) = [m]\frac{π}{3}+2πk[/m]
x_(2) = [m]\frac{5π}{3}+2πk[/m]
4)[m]tg^2x-tgx+1=0[/m]
сделаем замену переменной: tgx = t
[m]t^2-t+1=0[/m]
D = (-1)^(2)-4*1*1 = - 3
D < 0 => решений нет
5) а] arccos(1)+arcsin(0) = 0 + 0 = 0
б] [m]arccos(-\frac{1}{2})-arcsin(\frac{\sqrt{3}}{2}) = \frac{2π}{3} - \frac{π}{3} = \frac{π}{3}[/m]