Find the solution of \cos(2x)-\cos(x)=0 in the interval [0; 2\pi)

Talamancoeb 2021-12-17 Answered
Find the solution of cos(2x)cos(x)=0 in the interval [0;2π)
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nghodlokl
Answered 2021-12-18 Author has 33 answers
We remember that cos(2x)=cos2(x)sin2(x)
And we have
cos2(x)sin2(x)cos(x)=0
Now, get everything in terms of cosine:
sin2(x)+cos2(x)=1
sin2(x)=1cos2(x)
Therefore,
cos2(x)(1cos2(x))cos(x)=0
2cos2(x)cos(x)1=0
(cos(x)1)(2cos(x)+1)=0
Now, solve
cos(x)1=0
2cos(x)+1=0
cos(x)=1
Implies that x=0, as we're working in [0;2π). And it's the only value yielding one for cosine.
2cos(x)=1
cos(x)=12
Implies that x=2π3,4π3 as these values return 12 for cosine in the given interval.
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Ethan Sanders
Answered 2021-12-19 Author has 35 answers
cos2xcosx=02cos2xcosx1=0(2cosx+1)(cosx1)=0cosx=12 and cosx=1
cosx=12x=2π3;x=4π3
cosx=1x=0
2π is not included in given interval
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