2sin^2x+cos^2x-2=0 [-П;0]. Created by pdero. algebra-ru

2sin^2x+cos^2x-2=0 [-П;0]...

sedinalana 2(sin²x-1)+cos²x=0
-2cos²x+cos²x=0
-cos²x=0
cosx=0
x=π/2+πk,k∈z
-π≤π/2+πk≤0
-2≤1+2k≤0
-3≤2k≤-1
-3/2≤k≤-1/2
k=-1 x=π/2-π=-π/2

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19ton51 2sin²x+cos²x-2=0 sinx=-1 или sinx=1
sin²x+1-2=0 x= - π/2+2πn, n∈Z x=π/2+2πk, k∈Z
sin²x=1 x= ....,-5π/2, -π/2, 3π/2, ....... x= ....,-3π/2,π/2, 5π/2, ...
значит, в промежуток [-π;0] входят угол -π/2

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