Посчитать . интеграл от dx/(sinx+cosx)

A) 4cos^2(5x-π/3)=3⇒cos^2(5x-π/3)=3/4⇒cos(5x-π/3)=+√3/2 1. cos(5x-π/3)=√3/2⇒5x-π/3=+(√3/2)=+π/6+2πn⇒ 5x=π/6+2πn+π/3⇒5x=π/2+2πn⇒ x1=π/10+2πn/5 5x=-π/6+2πn+π/3⇒5x=π/6+2πn⇒ x2=π/30+2πn/5 2. cos(5x-π/3)=-√3/2⇒5x-π/3=+π-arccos(√3/2))=+π/6+2πn⇒ 5x=5π/6+2πn+π/3⇒5x=7π/6+2πn⇒ x3=7π/30+2πn/5 5x=-5π/6+2πn+π/3⇒5x=-π/2+2πn⇒ x4=-π/10+2πn/5 б) √sin^2(5x)=sin5x⇒isin5xi=sin5x 1)2πn< =5x< =π+2πn⇒sin5x=sin5x⇒ x∈[2πn/5; (π+2πn)/5] 2)π+2πn< 5x< 4πn⇒-sin5x=sin5x⇒2sin5x=0⇒sin5x=0⇒5x=πn⇒ 5x∉ (π+2πn; 4πn)⇒ в этом интервале решений нет 3) 2sinπ/4*cos3x = cos3x⇒cos3x(2*√2/2-1)=0⇒cos3x(√2-1)=0⇒cos3x=0⇒ 3x=π/2+πn⇒x=π/6+πn/3 4) 3sin^2(x) = cos ^2(x)+1⇒3*(1-cos^2( ^2(x)-1=0⇒ 4cos^2(x)=2⇒cos^2(x)=1/2⇒cosx=+√2/2⇒ x=+π/4+2πn x=+π/4+2πn

  b+20 при b=70   70+20=90при  b=23     23 + 20   =  43 при  b=11   11 + 20   =  31 14+bпри b=70       14 + 70   =  84 при  b=23     14 + 23   =  37 при  b=11     14 + 11   =  25 80-bпри b=70     80 − 70   =  10 при  b=23     80 − 23   =  57 при  b=11     80 − 11   =  69 b-9при b=70     70 − 9   =  61 при  b=23     23 − 9   =  14 при  b=11   11 − 9   =  2