How do you solve #cosx=3cosx-2#?

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Answer 1 · 2016-08-28T04:16:11

#x=0 or 2 pi#

If we let #cosx=w#, then we can simplify our equation:

#cosx=3cosx-2#
#w=3w-2#

Next, isolate the #w# terms to one side:

#w-w+2=3w-2-w+2#
#2=3w-w#
#2=2w#
#w=1#

Replace #w# with #cosx#:

#w=1#
#cosx=1#
#x=cos^-1(1)#
#x=0 or 2pi#

Note: Technically, since the #cosx# function completes a full cycle for every #2pi# radians, #x# could be more accurately defined as:

#x=2pin# where #ninZZ#