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evgenijdomrach
@evgenijdomrach
July 2022
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решить уравнение sin^2x+cos^22x=1
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Verified answer
Sin²x + cos²2x = 1
-1 + cos²2x + sin²x = 0
-sin²2x + sin²x = 0
-sin2x•sin2x + sin²x = 0
-4sin²x•cos²x + sin²x = 0
sin²x(-4cos²x + 1) = 0
sin²x = 0 и -4cos²x = -1
sinx = 0 и cos²x = 1/4
sinx = 0
x = πn, n ∈ Z
cosx = 1/2 и cosx = -1/2
x = ±π/3 + 2πn, n ∈ Z
и
x = ±2π/3 + 2πn, n ∈ Z.
5 votes
Thanks 14
evgenijdomrach
Большое спасибо! Самый короткий вариант решения.
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Answers & Comments
Verified answer
Sin²x + cos²2x = 1-1 + cos²2x + sin²x = 0
-sin²2x + sin²x = 0
-sin2x•sin2x + sin²x = 0
-4sin²x•cos²x + sin²x = 0
sin²x(-4cos²x + 1) = 0
sin²x = 0 и -4cos²x = -1
sinx = 0 и cos²x = 1/4
sinx = 0
x = πn, n ∈ Z
cosx = 1/2 и cosx = -1/2
x = ±π/3 + 2πn, n ∈ Z
и
x = ±2π/3 + 2πn, n ∈ Z.