sin²(π/8 + t) = sint + sin²(π/8 - t)
(1 - cos(π/4 + 2t))/2 = sint + (1 - cos(π/4 - 2t))/2
1 - ( (√2/2)•cos2t - (√2/2)•sin2t ) = 2sint + 1 - ( (√2/2)•cos2t + (√2/2)•sin2t )
1 - (√2/2)•cos2t + (√2/2)•sin2t = 2sint + 1 - (√2/2)•cos2t - (√2/2)•sin2t
2sint - √2sin2t = 0
2sint - 2√2•sint•cost = 0
2sint•( 1 - √2•cost) = 0
ответ: πn, n ∈ Z ; ± π/4 + 2πk, k ∈ Z
sin²(π/8 + t) = sint + sin²(π/8 - t)
sin²x = (1 - cos2x)/2(1 - cos(π/4 + 2t))/2 = sint + (1 - cos(π/4 - 2t))/2
cos(α + β) = cosα•cosβ - sinα•sinβ - косинус суммыcos(α - β) = cosα•cosβ + sinα•sinβ - косинус разности1 - ( (√2/2)•cos2t - (√2/2)•sin2t ) = 2sint + 1 - ( (√2/2)•cos2t + (√2/2)•sin2t )
1 - (√2/2)•cos2t + (√2/2)•sin2t = 2sint + 1 - (√2/2)•cos2t - (√2/2)•sin2t
2sint - √2sin2t = 0
sin2x = 2•sinx•cosx - синус двойного аргумента2sint - 2√2•sint•cost = 0
2sint•( 1 - √2•cost) = 0
sint = 0 ⇔ t = πn, n ∈ Z1 - √2•cost = 0 ⇔ cost = 1/√2 ⇔ t = ± π/4 + 2πk, k ∈ Zответ: πn, n ∈ Z ; ± π/4 + 2πk, k ∈ Z