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vb0988650556nb
@vb0988650556nb
July 2022
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розв'яжіть рівняння корінь cos2x- tg2x=1
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армения20171
Cos2x-tg2x=1
(1-tg²x)/(1+tg²x)-2tgx/(1-tg²x)=1
(1-tg²x)²-2tgx(1+tg²x)=(1+tg²x)(1-tg²x)
1-2tg²x+tg⁴x-2tgx-2tg³x=1-tg⁴x
2tg⁴x-2tg³x-2tg²x-2tgx=0
2tgx(tg³x-tg²x-tgx-1)=0
1)tgx=0
x=πk;k€Z
2)tg³x-tg²x-tgx-1=0
tgx(tg²x-1)-(tg²x+1)=0
1+tg²x#0;1-tg²x#0
-tgx(1-tg²x)-(1+tg²x)=0
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Answers & Comments
(1-tg²x)/(1+tg²x)-2tgx/(1-tg²x)=1
(1-tg²x)²-2tgx(1+tg²x)=(1+tg²x)(1-tg²x)
1-2tg²x+tg⁴x-2tgx-2tg³x=1-tg⁴x
2tg⁴x-2tg³x-2tg²x-2tgx=0
2tgx(tg³x-tg²x-tgx-1)=0
1)tgx=0
x=πk;k€Z
2)tg³x-tg²x-tgx-1=0
tgx(tg²x-1)-(tg²x+1)=0
1+tg²x#0;1-tg²x#0
-tgx(1-tg²x)-(1+tg²x)=0