Ответы на вопрос:
12(сosx -cos²x/2 - sin²x/2)² = 10 - 13cosx
12(cosx -(cos²x/2 +sin²x/2))² = 10 - 13cosx
12(cosx -1)² = 10 -13cosx
12(cos²x -2cosx +1) = 10 -13cosx
12cos²x -24cosx +12 - 10 +13cosx = 0
12cos²x -11cosx +2 = 0
cosx = t
12t² -11t +2 = 0
d = 121 - 96 = 25
t₁ = 16/24= 2/3 t₂ = 1/4
сosx = 2/3 cosx = 1/4
x = +- arccos2/3 + 2πk , k ∈z x = =-arccos1/4 + 2πn , n ∈z
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