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danechkamich
@danechkamich
August 2022
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решите, пожалуйста, уравнение. 2cos2x+cos^2(x/2) -10cos(5п/2 - x)+7/2=1/2 *cosx Спасибо.
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manyny06
Verified answer
2-4sin²x+1\2+1\2cosx-10sinx+7\2=1\2cosx
2sin²x+5sinx-3=0
sinx=t≤1
2t²+5t-3=0
D=25-4*2*(-3)= 49 √D=7
t₁=(-5+7)\4 = 1\2 t₂=(-5-7)\4= -3 - ∉ по ОДЗ
sinx=1\2 ⇒ x=(-1) в степени к * π\6 + πk. k∈Z
1 votes
Thanks 9
Радость25
Полное решение смотри во вложении:
1 votes
Thanks 8
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Answers & Comments
Verified answer
2-4sin²x+1\2+1\2cosx-10sinx+7\2=1\2cosx2sin²x+5sinx-3=0
sinx=t≤1
2t²+5t-3=0
D=25-4*2*(-3)= 49 √D=7
t₁=(-5+7)\4 = 1\2 t₂=(-5-7)\4= -3 - ∉ по ОДЗ
sinx=1\2 ⇒ x=(-1) в степени к * π\6 + πk. k∈Z