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