Решение
1) 2sin^2x - sinx - 1=0
D = 1 + 4*2*1 = 25
a) sinx = (1-3)/4
sinx = -1/2
x = (-1)^n*arcsin(-1/2) + πn, n∈Z
x = (1)^(n+1)*arcsin(1/2) + πn, n∈Z
x1 = (1)^(n+1*(π/6) + πn, n∈Z
b) sinx = (1+3)/4
sinx = 1
x2 = π/2 + 2πk,k ∈Z
Ответ: x1 = (1)^(n+1)*(π/6) + πn, n∈Z; x2 = π/2 + 2πk,k ∈Z
2tg^2x + 3tgx - 2=0
D = 9 + 4*2*2 = 25
a) tgx = (-3 -5)/4
tgx = -2
x1 = arctg(-2) + πn, n∈Z
x1 = - arctg(2) + πn, n∈Z
b) tgx = (-3+5)/4
tgx = 1/2
x2 = arctg(1/2) + πk, k∈Z
Ответ: x1 = - arctg(2) + πn, n∈Z; x2 = arctg(1/2) + πk, k∈Z.