Ответы на вопрос:
2sin^2x + 2sinx - 1= 0 2sin^2x + 2sinx =1 2sin^2x + 2sinx = sin^2x +cos^2x 2(1-cos^2x) + 2sinx =sin^2x +cos^2x 2-2cos^2x + 2sinx = sin^2x +cos^2x 2+2sinx = sin^2x + 3cos^2x 3+2sinx = sin^2x + 3cos^2x +1 3-3cos^2x=sin^2x - 2sinx +1 3(1-cos^2x)= (1-sinx)^2 3*sin^2x = (1-sinx)^2 √3sinx=1-sinx √3sinx + sinx=1 sinx(√3+1)=1 sinx=1/√3+1
b1=v6 b2=v18 q=b2/b1=v3
s4=b1(q^4-1)/q-1={v6(9-1)}/(v3 -1)= 8v6(v3+1)/{(v3-1)(v3+1)=8v6(v3+1)/2=4v6(v3+1)=
=4v18-4v6=12v2-4v6
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