НУЖНО ПОДРОБНОЕ РЕШЕНИЕ В2 и В3

Question

Математика: НУЖНО ПОДРОБНОЕ РЕШЕНИЕ В2 и В3

mirochkanervnya · Accepted Answer

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=+(-)arccos(√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))=+(-)5π/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(x))-cos ^2(x)-1=0⇒
4cos^2(x)=2⇒cos^2(x)=1/2⇒cosx=+(-)√2/2⇒
x=+(-)π/4+2πn
x=+(-)3π/4+2πn

MOGZ ответил