Если правильно понял формулу:
b^{2}*20rb−20r*(20r)^{2}*b=20rb(b^{2}-(20r)^{2})=20rb(b-20r)(b+20r)
подставим r=1/20 ,b=−2 в "20rb(b-20r)(b+20r)", получим:
20*1/20*(-2)*(-2-20*1/20)(-2+20*1/20)=(-2)*(-2-1)*(-2+1)=(-2)*(-3)*(-1)=-6;
Ответ: -6
1/sinx + 1/cos(7π/2 + x)=2
1/sinx + 1/cos(3π/2+2π+x)=2
1/sinx +1/cos(3π/2+x)=2
1/snx + 1/cos(π/2+π+x)=2
1/sinx + 1/(-cos(π/2+x))=2
1/sinx +1/sinx=2
2/sinx=2sinx | *(1/2 *sinx);sinx≠0
sin^2 x=1
|sinx|=1
sinx=-1 ili sinx=1
x=-π/2+2πn x=π/2+2πn
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x⊂[-5π/2; -π]
-5π/2 ≤-π/2+2πn≤-π -5π/2≤π/2+2πn≤-π
-5π/2+π/2≤2πn≤-π+π/2 -3π/(2π)≤n≤ -π/(2π)
-4π/2≤2πn≤-π/2 -1,5≤n≤ -1/2 ; n-celoe
(-2π)/(2π)≤n≤-π/(2*2π); n=-1
-1≤n≤-1/4 x=π/2-2π; x=-3π/4
n=-1 -----------
n=-1; -π/2-2π=-5π/2
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x>0 x^2-3x-40=0 x=(3+13)/2=8
4x²+12x+9-4(1-2x+x²)=1
4x²+12x+9-4+8x-4x²=1
20x+5=1
20x=1-5
20x=-4
x=-0.2