-1 + cos²x +cos²x -cosx =0 .
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2cos²x -cosx -1 =0 ; замена : t = cosx, где -1 ≤ t ≤ 1 .
2t ² - t -1 = 0 ; * * * D =1² - 4*2*(-1) = 9 =3² * * *
[ t = (1- 3) / (2*2) ; t = (1+3) / (2*2) .⇔[ t = -1 / 2) ; t = 1.
[ cosx = -1 / 2 ; cosx = 1 ⇒ [ x =±2π/3 + 2πn ; x = 2πn , n ∈Z .
ответ: { ±2π / 3 + 2πn ; x = 2πn , n ∈Z } .
* * * * * * *
cosx =a , где - 1 ≤ a ≤ 1 . ⇒ x = ±arccosa + 2πn , n ∈ Z .
arccos(-1/2) = π -π/3 =2π / 3