How do you find exact solutions of #cos2x - cosx = 0# in the interval [0,2π) ?

1 Answer
VNVDVI
Mar 24, 2018

#x=0, (2pi)/3, (4pi)/3#

Explanation:

Recall that

#cos(2x)=cos^2x-sin^2x#.

Now, we have

#cos^2x-sin^2x-cosx=0#

However, we want our equation in terms of only one trigonometric function. We can easily get everything in terms of cosine:

#sin^2x+cos^2x=1#

#sin^2x=1-cos^2x#

Thus,

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

#2cos^2x-cosx-1=0#

This resembles a quadratic function, except it is composed of cosines. So, let's factor it as we would any quadratic function. I'll leave that part up to you, as it's just a matter of factoring as you would factor any other quadratic however you like:

#(cosx-1)(2cosx+1)=0#

Now, we solve the following:

#cosx-1=0#
#2cos+1=0#

#cosx=1#

Implies that #x=0# since we're working in #[0, 2pi).# It's the only value yielding one for cosine.

#2cosx=-1#
#cosx=-1/2#

Implies that #x=(2pi)/3, (4pi)/3# as these two values return #-1/2# for cosine in the interval #[0, 2pi)#

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