Solve the trigonometric equation #3sinx -5cosx=2# ?

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Answer 1 · 2018-07-11T18:50:49

#x=pi-arcsin(sqrt(2/17))+arcsin(5/sqrt(34))+2kpi# or #x=arcsin(sqrt(2/17))+arcsin(5/sqrt(34))+2kpi#

We have to solve

#3sin(x)-5cos(x)=2#
note that #3^2+5^2=9+25=34#

so we can write

#3/sqrt(34)sin(x)-5/sqrt(34)cos(x)=2/sqrt(34)# so exist

#cos(phi)=3/sqrt(34)#

#sin(phi)=5/sqrt(34)#
so we get

#sin(x)cos(phi)-sin(phi)cos(x)=2/sqrt(34)#
#sin(x-phi)=2/sqrt(34)#

#x=pi-arcsin(sqrt(2/17))+arcsin(5/sqrt(34))+2kpi#

#x=arcsin(sqrt(2/17))+arcsin(5/sqrt(34))+2kpi#
Note that #2/sqrt(34)=sqrt(4/34)=sqrt(2/17)#