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nikit122
@nikit122
July 2022
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[tex]cos2x-sin^{2} x=0,25 [/tex]
Решите,пожалуйста
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rasimagaliullov
Косинус двойного угла
cos2α=cos²α-sin²α
sin
²
x+cos²x=1
(основное тригонометрическое тождество),
cos²x=1-sin²x
cos
²x-sin²x
-sin²x=0.25
1-sin²x-2sin²x=0.25
-3sin²x=-0.75
sin²x=0,25
sinx=0.5 sinx=-0.5
sinx=1/2 sinx=-1/2
x=π/6 x=-π/6
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Answers & Comments
sin²x+cos²x=1 (основное тригонометрическое тождество),
cos²x=1-sin²x
cos²x-sin²x-sin²x=0.25
1-sin²x-2sin²x=0.25
-3sin²x=-0.75
sin²x=0,25
sinx=0.5 sinx=-0.5
sinx=1/2 sinx=-1/2
x=π/6 x=-π/6