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sustinianvanxh
@sustinianvanxh
July 2022
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6
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sin^4(cos^3(3x)) + cos^4(cos^3 (3x)) =1
Найдите все x на промежутке [0;1]
99 балов
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Segrif
Sin^2(cos^3(3x)) = t, 0 <= t <= sin^2(1) < 1
cos^2(cos^3(3x)) = 1 - t
t^2 + (1-t)^2 = 1
t^2 + t^2 - 2t + 1 = 1
t(t - 1) = 0
t = 1 - не подходит
t = 0
sin(cos^3(3x)) = 0
cos^3(3x) = y, |y| <= 1
sin(y) = 0
y = Пk, П > 1 -> y = 0
cos^3(3x) = 0
cos(3x) = 0
3x = П/2 + Пk
x = П/6 + Пk/3
0 <= x <= 1
0 < П/6 < 1
П/3 > 1
x = П/6 - единственное решение на [0;1]
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Answers & Comments
cos^2(cos^3(3x)) = 1 - t
t^2 + (1-t)^2 = 1
t^2 + t^2 - 2t + 1 = 1
t(t - 1) = 0
t = 1 - не подходит
t = 0
sin(cos^3(3x)) = 0
cos^3(3x) = y, |y| <= 1
sin(y) = 0
y = Пk, П > 1 -> y = 0
cos^3(3x) = 0
cos(3x) = 0
3x = П/2 + Пk
x = П/6 + Пk/3
0 <= x <= 1
0 < П/6 < 1
П/3 > 1
x = П/6 - единственное решение на [0;1]