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lanaromanova9
@lanaromanova9
August 2022
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2sin^2x-7sin2x=16cos^2x Решите пожалуйста ❤️
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sedinalana
Verified answer
Разделим на cos²x
2tg²x-14tgx-16=0
tgx=a
a²-7a-8=0
a1+a2=7 U a1*a2=-8
a1=-1⇒tgx=-1⇒x=-π/4+πn,n∈z
a2=8⇒tgx=8⇒x=arctg8+πn,n∈z
6 votes
Thanks 9
lanaromanova9
там (-7sin*2x) если мы делим на cos^2x не может получится tgx
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Answers & Comments
Verified answer
Разделим на cos²x2tg²x-14tgx-16=0
tgx=a
a²-7a-8=0
a1+a2=7 U a1*a2=-8
a1=-1⇒tgx=-1⇒x=-π/4+πn,n∈z
a2=8⇒tgx=8⇒x=arctg8+πn,n∈z