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annagrishchenko
@annagrishchenko
August 2022
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2sin^2x+5sinxcosx-7cos^2x=0
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m11m
Verified answer
Делим все на cos²x:
(2sin²x/cos²x)+(5sinxcosx/cos²x) -(7cos²x/cos²x)=0/cos²x
2tg²x+5tgx-7=0
y=tgx
2y²+5y-7=0
D=5² -4*2*(-7)=25+56=81
y₁=(-5-9)/4= -3.5
y₂=(-5+9)/4=1
При у= -3,5
tgx= -3.5
x= -arctg3.5 +пk, k∈Z
При у=1
x=п/4 + пk, k∈Z
Ответ: -arctg3.5 + пk, k∈Z;
п/4 + пk, k∈Z.
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Answers & Comments
Verified answer
Делим все на cos²x:(2sin²x/cos²x)+(5sinxcosx/cos²x) -(7cos²x/cos²x)=0/cos²x
2tg²x+5tgx-7=0
y=tgx
2y²+5y-7=0
D=5² -4*2*(-7)=25+56=81
y₁=(-5-9)/4= -3.5
y₂=(-5+9)/4=1
При у= -3,5
tgx= -3.5
x= -arctg3.5 +пk, k∈Z
При у=1
x=п/4 + пk, k∈Z
Ответ: -arctg3.5 + пk, k∈Z;
п/4 + пk, k∈Z.