Решие
<span>sin α/2 =2/√7 , 0<α<п
sin</span>²(α/2) = (1 - cosα)/2
(2/√7)² = (1 - cosα)/2
1 - cosα = 8/7
cosα = 1 - 8/7
cosα = - 1/7
sinα = √(1 - cos²α) = √(1 - (- 1/7)²) = √(1 - 1/49) = √(48/49) =
= 4√3 / 7
1) (7x-5)²-(2x+1)²=0
(7x-5-2x-1)*(7x-5+2x+1)=0
(5x-6)*(9x-4)=0
x1= 6/5
x2= 4/9
2) a²(4a-3)-(4a-3)=0
(4a-3)(a²-1)=0
(4a-3)(a-1)(a+1)=0
a1= 4/3
a2= 1
a3= -1