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deevil
@deevil
August 2022
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sedinalana
2/(tg²x+1)=sin2x
2:1/cos²x=sin2x
2cos²x-2sinxcosx=0
2cosx(cosx-sinx)=0
cosx=0
x=π/2+πk,k∈z
cosx-sinx=0/cosx
1-tgx=0
tgx=1
x=π/4+πk,k∈z
---------------------------------
-9π/2≤π/2+πkπ-3π/2
-9≤1+2k≤-3
-10≤2k≤-4
-5≤k≤-2
k=-5⇒x=π/2-5π=-9π/2
k=-4⇒x=π/2-4π=-7π/2
k=-3⇒x=π/2-3π=-5π/2
k=-2⇒x=π/2-2π=-3π/2
-9π/2≤π/4+πk≤-3π/2
-18≤1+4k≤-6
-19≤4k≤-5
-19/4≤k≤-5/4
k=-4⇒x=π/4-4π=-15π/4
k=-3⇒x=π/4-3π=-11π/4
k=-2⇒x=π/4-2π=-7π/4
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Answers & Comments
2:1/cos²x=sin2x
2cos²x-2sinxcosx=0
2cosx(cosx-sinx)=0
cosx=0
x=π/2+πk,k∈z
cosx-sinx=0/cosx
1-tgx=0
tgx=1
x=π/4+πk,k∈z
---------------------------------
-9π/2≤π/2+πkπ-3π/2
-9≤1+2k≤-3
-10≤2k≤-4
-5≤k≤-2
k=-5⇒x=π/2-5π=-9π/2
k=-4⇒x=π/2-4π=-7π/2
k=-3⇒x=π/2-3π=-5π/2
k=-2⇒x=π/2-2π=-3π/2
-9π/2≤π/4+πk≤-3π/2
-18≤1+4k≤-6
-19≤4k≤-5
-19/4≤k≤-5/4
k=-4⇒x=π/4-4π=-15π/4
k=-3⇒x=π/4-3π=-11π/4
k=-2⇒x=π/4-2π=-7π/4