Cos([latex] \frac{ \pi }{2} +x)=1[/latex][latex]log_{5} (5-x)= log_{5} 3[/latex][latex] \sqrt{2} cosx-1=0[/latex]

Question

Cos([latex] \frac{ \pi }{2} +x)=1[/latex][latex]log_{5} (5-x)= log_{5} 3[/latex][latex] \sqrt{2} cosx-1=0[/latex]

Математика

Cos([latex] \frac{ \pi }{2} +x)=1[/latex] [latex]log_{5} (5-x)= log_{5} 3[/latex] [latex] \sqrt{2} cosx-1=0[/latex]

Гость

Ответ(ы) на вопрос:

Accepted Answer

cos(π/2+x) = 1 π/2 + x = 2πk ; k∈Z x= 2πk - π/2 = (4k-1)·π/2 ; k∈Z log(5) (5-x) - log(5) 3 = 0 log(5) [(5-x)/3] = 0 (5-x)/3 = 1 5-x = 3 x=2 cosx = 1/√2 = √2/2 x = +/-π/4 + 2πk ; k∈Z

Cos([latex] \frac{ \pi }{2} +x)=1[/latex]﻿﻿[latex]log_{5} (5-x)= log_{5} 3[/latex]﻿[latex] \sqrt{2} cosx-1=0[/latex]

Cos([latex] \frac{ \pi }{2} +x)=1[/latex][latex]log_{5} (5-x)= log_{5} 3[/latex][latex] \sqrt{2} cosx-1=0[/latex]