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Ярррик898
@Ярррик898
July 2022
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Тригонометрические уравнения
А) 37cos x + 5sin^2 x + 5cos^2 x=4
Б) cos^2 2x/37=3/4
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Dимасuk
Verified answer
А) 37cosx + 5sin²x + 5cos²x = 4
37cosx + 5 = 4
37cosx = -1
cosx = -1/37
x = ±arccos(-1/37) + 2πn, n ∈ Z
cos²(2x/37) = 3/4
cos(2x/37) = ±√3/2
cos(2x/37) = √3/2
2x/37 = ±π/6 + 2πn, n ∈ Z
x = ±37π/12 + 37πn, n ∈ Z
cos(2x/37) = -√3/2
2x/37 = ±5π/6 + 2πn, n ∈ Z
x = ±185π/3 + 37πn, n ∈ Z.
1 votes
Thanks 2
Ярррик898
А как ты избавился от синуса и косинуса сразу
Поясни пж
Dимасuk
5sin²x + 5cos²x = 5(sin²x + cos²x) = 5•1 = 5. Это основное тригонометрическое тождество.
Dимасuk
sin²a + cos²a = 1.
Ярррик898
Спасибо большое
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Answers & Comments
Verified answer
А) 37cosx + 5sin²x + 5cos²x = 437cosx + 5 = 4
37cosx = -1
cosx = -1/37
x = ±arccos(-1/37) + 2πn, n ∈ Z
cos²(2x/37) = 3/4
cos(2x/37) = ±√3/2
cos(2x/37) = √3/2
2x/37 = ±π/6 + 2πn, n ∈ Z
x = ±37π/12 + 37πn, n ∈ Z
cos(2x/37) = -√3/2
2x/37 = ±5π/6 + 2πn, n ∈ Z
x = ±185π/3 + 37πn, n ∈ Z.
Поясни пж