How do you solve #7cosx=5cosx+sqrt3# for #0<=x<=2pi#?

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Answer 1 · 2016-11-17T09:46:11

#x=pi/6# or #(11pi)/6#

As #7cosx=5cosx+sqrt3#, transposing #5cosx# to LHS we get

#7cosx-5cosx=sqrt3#

or #2cosx=sqrt3#

or #cosx=sqrt3/2#

In the range #0<=x<=2pi#, we can have #cosx=sqrt3/2#

when #x=pi/6# or #x=2pi-pi/6=(11pi)/6#

Hence, #x=pi/6# or #(11pi)/6#