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

1 Answer
VNVDVI
Mar 24, 2018

x=0, (2pi)/3, (4pi)/3x=0,2π3,4π3

Explanation:

Recall that

cos(2x)=cos^2x-sin^2xcos(2x)=cos2x−sin2x.

Now, we have

cos^2x-sin^2x-cosx=0cos2x−sin2x−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=1sin2x+cos2x=1

sin^2x=1-cos^2xsin2x=1−cos2x

Thus,

cos^2x-(1-cos^2x)-cosx=0cos2x−(1−cos2x)−cosx=0

2cos^2x-cosx-1=02cos2x−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(cosx−1)(2cosx+1)=0

Now, we solve the following:

cosx-1=0cosx−1=0
2cos+1=02cos+1=0

cosx=1cosx=1

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

2cosx=-12cosx=−1
cosx=-1/2cosx=−12

Implies that x=(2pi)/3, (4pi)/3x=2π3,4π3 as these two values return -1/2−12 for cosine in the interval [0, 2pi)[0,2π)

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