Question

1.Решить уравнения:
а)sin x/2 = 1/2
б)2cos^2x - sin x - 1 = 0
в)2cos^2x - sin x - 1 = 0

2.Найти cos a, если sina =√3/2 и п/4

Answer 1

A) x/2 = (-1)^k*arcsin(1/2) + πk
x= ((-1)^k*(π/6) + πk)/2
B) 2*cos^2x - sinx - 1 = 0
2*((1+cos2x)/2) - sinx - 1 = 0
Раскроем скобки
1 + cos2x - sinx - 1 = 0
Cos2x= 1 - 2sin^2x
1 - 2sin^2x - sinx = 0
Sinx = t
2t^2 + t - 1 = 0
t1 = -1 t2 = 1/2
Sinx = -1 sinx = 1/2
X1 = (-1)^k*(3π/2) + πk
X2 = (-1)^k*(π/6) + πk