How do you solve for x in #2cos(3x-1)=0#?

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Answer 1 · 2015-02-12T11:53:00

If #2cos(3x-1)=0# then so is #cos(3x-1)#

If a #cos theta=0# then #theta=90^0# plus any multiple of #180^0#
(the cosine function crosses the #x#-axis twice every period)

So we have
#3x-1=90^0+k*180^0#

#->3x=91^0+k*180^0#

#->x=91^0/3+k*180^0/3#

Answer:
#x=(30 1/3) ^0 +k*60^0#

Extra:
If you work in radians (you didn't say), then solving would go along the same lines, only then you would work with

#3x-1=pi/2 +k.pi#