Помогите срочно! !!!!!!!. Created by deevil. matematika-ru

Помогите срочно! !!!!!!!...

sedinalana Tg²x+1=1/cos²x
2/(tg²x+1)=2cos²x
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+πk≤-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≤-7
-19/4≤k≤-7/4
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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MrCalling $\dfrac{2}{tg^2x+1}=sin2x\\ 2cos^2x-2sinxcosx=0\\2cosx(cosx-sinx)=0\\cosx=0\\x= \dfrac{\pi}{2}+\pi k,k\in Z\\cosx-sinx=0\\tgx=1\\x= \dfrac{\pi}{4}+\pi k,k\in Z$
рассмотрим корни
-9p/2≤p/2+pk≤-3p/2
-9≤1+2k≤-3
-10≤2k≤-4
-5≤k≤-2
k=-5, x=p/2-5pk=-9p/2
k=-4,x=-7p/2
k=-3,x=-5p/2
k=-2,x=-3p/2
-9p/2≤p/4+pk≤-3p/2
-9*2≤1+2*2k≤-2*3
-18≤1+4k≤-6
-19≤4k≤-7
-4,5≤k≤1,75
k∈[-4;-2]
k=-4,x=p/4-4p=-15p/4
k=-3,x=p/4-3p=-11p/4
k=-2,x=p/4-2p=-7p/4

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