Помогите!!! 6 sin x=3-8 cos^2x. Created by Ameya. algebra-ru

Помогите!!! 6 sin x=3-8 cos^2x...

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6sinx-3+8-8sin²x=0
sinx=a
8a²-6a-5=0
D=36+160=196
a1=(6-14)/16=-1/2⇒sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn,n∈z
a2=(6+14)/16=5/4⇒sinx=5/4>1 нет решения

2 votes Thanks 3

oganesbagoyan

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6 sin x=3-8 cos²x ;
3 - 8 (1-sin²x ) - 6sinx =0 ;
8sin²x -6sinx -5 =0 ; * * * -1 ≤ sinx ≤ 1 * * *
[ sinx =(3 - 7)/8 ; sinx = (3 + 7)/8 .⇔ [ sinx = -1/2 ; sinx = 1,25 >1.
sinx = -1/2 ⇒ x =(-1) ^(n+1)*π/6 + π*n , n∈Z.

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reshite trigonometricheskoe uravnenie78de5f7c1729181cb29df756372383cc 78486

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koren iz 28 koren iz 7 koren iz 7

ustanovite sootvetstvie mezhdu priznakom energeticheskogo obmena i ego etapom priz

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