Natali Belska
Просветленный
(48671)
1 неделю назад
2^(sin^2 x) + 2^(cos^2 x) = 3
2^(sin^2 x) + 2^(1 - sin^2 x) = 3
2^(sin^2 x) + 2/ (2^sin^2 x) = 3 ---> sin^2 x = t
t^2 - 3t + 2 = 0
t(1,2) = [3 + - V(3^2 - 4*2)]/2 = (3 + - 1)/2
t1 = (3+1)/2 = 2
t2 = (3-1)/2 = 1
sin^2 x = t1 = 2 - лишний корень (так как -1 < sin x < 1) =>
sin^2 x = t2 = 1
sin x = - 1 => x1 = ...
и sin x = 1 => x2 = ...