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together43
@together43
October 2021
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Решите неравенство Tg^2x<1 (tg квадрат x)
подробно, с рисунком и ответом.
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sedinalana
Verified answer
(1-cos2x)/(1+cos2x)<1
(1-cos2x)/(1+cos2x)-1<0
(1-cos2x-1-cos2x)/(1+cos2x)<0
2cos2x/(1+cos2x)>0
1)cos2x>0 U cos2x>-1⇒cos2x>0⇒2x∈(-π/2+2πn;π/2+2πn,n∈z)⇒
x∈(-π/4+πn;π/4+πn,n∈z)
2)cos2x<0 U cos2x<-1⇒cos2x<-1 нет решения
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Answers & Comments
Verified answer
(1-cos2x)/(1+cos2x)<1(1-cos2x)/(1+cos2x)-1<0
(1-cos2x-1-cos2x)/(1+cos2x)<0
2cos2x/(1+cos2x)>0
1)cos2x>0 U cos2x>-1⇒cos2x>0⇒2x∈(-π/2+2πn;π/2+2πn,n∈z)⇒
x∈(-π/4+πn;π/4+πn,n∈z)
2)cos2x<0 U cos2x<-1⇒cos2x<-1 нет решения