How do you solve #4cos^2x-1=0#?

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Answer 1 · 2016-10-21T01:05:00

# x=+-(4pi)/3,+-(2pi)/3,+-(5pi)/3,+-(pi)/3 #

This is a trivial quadratic equation in #cosx#

# 4cos^2x-1=0 #
# :. 4cos^2x = 1 #
# :. cos^2x = 1/4 #
# :. cosx = +-1/2 #

In order to solve this you need to be familiar with the graphs of the trig functions along with common values:

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The range of soutions is not specified in the question, so lets assume we want a solutio in radians, and #-2pi <= x <= 2pi#

Consider, # cos=1/2 => x=-2pi+pi/3,-pi/3,pi/3, 2pi-pi/3#
# :. x=(-5pi)/3,(-pi)/3,pi/3, (5pi)/3#

Consider, # cos=-1/2 => x=-2pi+(2pi)/3,(-2pi)/3,(2pi)/3, 2pi-(2pi)/3#
# :. x=(-4pi)/3,(-2pi)/3,(2pi)/3, (4pi)/3#

So the solutions are:
# :. x=+-(4pi)/3,+-(2pi)/3,+-(5pi)/3,+-(pi)/3 #