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

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