сколько корней имеет уравнение sin^2(2x) = 1/4 в промежутке {-П/2, П/2}. Created by elnur4523. algebra-ru

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сколько корней имеет уравнение
sin^2(2x) = 1/4 в промежутке {-П/2, П/2}

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sedinalana

Verified answer

(1-cos4x)/2=1/4
1-cos4x=1/2
cos4x=1/2
4x=-π/3+2πn U 4x=π/3+2πk
x=-π/12+πn/2,n∈z U x=π/12+πk/2,k∈z
-π/2≤-π/12+πn/2≤π/2
-6≤-1+6n≤6
-5≤6n≤7
-5/6≤n≤7/6
n=0⇒x=-π/12
n=1⇒x=-π/12+π/2=5π/12
-π/2≤π/12+πk/2≤π/2
-6≤1+6k≤6
-7≤6k≤5
-7/6≤k≤5/6
k=-1⇒x=π/12-π/2=-5π/12
k=0⇒x=π/12

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