Question

Решите пожалуйста: 8sin^4x+10sin^x-3=0

Answer 1

Решение
8sin⁴x + 10sin²x - 3 = 0
sin²x = t, t ≥ 0
8t² + 10t - 3 = 0
D = 100 + 4*8*3 = 196
t₁ = (- 10 - 14)/16
t₁ = - 24/16
t₁ = - 1,5, не удовлетворяет условию  t ≥ 0
t₂ = (- 10 + 14)/16
t₂ = 4/16
t₂ = 1/4
sin²x = 1/4
sinx = - 1/2
sinx = 1/2
1) sinx = - 1/2
x = (-1)^n*arcsin(-1/2) + πn, n∈Z
x = (-1)^(n+1)*arcsin(1/2) + πn, n∈Z
x = (-1)^(n+1)*arcsin(1/2) + πn, n∈Z
x = (-1)^(n+1)*(π/6) + πn, n∈Z
2)  sinx = 1/2
x = (-1)^k*arcsin(1/2) + πk, n∈Z
x = (-1)^k*(π/6) + πk, k∈Z