Найдите члены прогрессии (гометр) ели а) b1=3 q=2 sn=189 б) b1=3 q=одна третья 1/3 sn=4целых13/27

a) sn=b1*(q(в степени)n-1)/(q-1)

      189=3*(2(в степени)n-1)/1

      63=2(в степени)n-1

      64=2(в степени)n

      n=6

б) sn=b1*(q(в степени)n-1)/(q-1)

      4целых13/27=3*(1/3(в степени)n-1)/-2/3

     

       

     

1) sin2x >/ 0

arcsin(0) + 2πk </ 2x </ π-arcsin(0) + 2πk , k € Z

0 + 2πk </ 2x </ π-0 + 2πk , k € Z

2πk </ 2x </ π + 2πk , k € Z / : 2

πk </ x </ π/2 + πk , k € Z

2) cos x/2 < √2/2

arccos(√2/2) + 2πk < x/2 < 2π - arccos(√2/2) + 2πk , k € Z

π/4 + 2πk < x/2 < 2π - π/4 + 2πk , k € Z

π/4 + 2πk < x/2 < 8π/4 - π/4 + 2πk , k € Z

π/4 + 2πk < x/2 < 7π/4 + 2πk , k € Z / : 1/2

π/2 + 4πk < x < 7π/2 + 4πk , k € Z

3) tg ( x - π/3 ) > √3

Замена (x-π/3) = a

tg a > √3

arctg(√3) + πk > a > π/2 + πk , k € Z

π/3 + πk > x - π/3 > π/2 + πk , k € Z

π/3 + π/3 + πk > x > π/2 + π/3 + πk , k € Z

2π/3 + πk > x > 3π/6 + 2π/6 + πk , k € Z

2π/3 + πk > x > 5π/6 + πk , k € Z

4) ctg (x+π/3) </ - 1/√3

Замена (x+π/3) = a

ctg a </ - 1/√3

arcctg(-1/√3) + πk </ a </ π + πk , k € Z

(π-arcctg(1/√3)) + πk </ a </ π + πk , k € Z

(π-π/3) + πk </ a </ π + πk , k € Z

2π/3 + πk </ a </ π + πk , k € Z

2π/3 + πk </ x+π/3 </ π + πk , k € Z

2π/3 - π/3 + πk </ x </ π - π/3 + πk , k € Z

π/3 + πk </ x </ 2π/3 + πk , k € Z

1(cos²2t-sin²2t)(cos²2t+sin²2t)=cos²2t-sin²2t=cos4t 2 sina=-√(1-cos²a)=-√(1-225/289)=-√(64/289)=-8/17 sin2a=2sinacosa=2*(-8/17)*15/17=-240/289 cos2a=cos²a-sin²a=225/289-64/289=161/289 tg2a=sin2a/cos2a=-240/289: 161/289=-240/289*289/161=-240/161 3 (sin3acos2b+cos3asin2b-sin3acos2b+cos3asin2b)/(cos3acos2b-sin3asin2b+ +cos3acos2b+sin3asin2b)=2cos3asin2b/2cos3acos2b=sin2b/cos2b=tg2b 4 sina=-√(1-cos²a)=-√(1-4/9)=-√5/3 cosb=-√(1-sin²b)=-√(1-1/9)=-2√2/3 sin2a=2sinacosa=2*(-√5/3)*2/3=-4√5/9 sin2b=2sinbcosb=2*1/3*(-2√2/3)=-4√2/9 cos2a=cos²a-sin²a=4/9-5/9=-1/9 cos2b=cos²b-sin²b=8/9-1/9=7/9 sin(2a+2b)=sin2acos2b+cos2asin2b=-4√5/9*7/9-1/9*(-4√2/9)= =-28√5/81+4√2/81=4(√2-7√5)/81