Решение
8sin⁴<span>x + 10sin</span>²<span>x - 3 = 0
</span>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
1)0,5*10-3 =2;
2)-10+9,7= -0,3