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Find all solutions to the equation in the interval [0, 2π). (3 points) cos 4x - cos 2x = 0 0, two pi divided by three. , four pi divided by three. pi divided by six , pi divided by two , five pi divided by six , seven pi divided by six , three pi divided by two , eleven pi divided by six 0, pi divided by three. , two pi divided by three. , π, four pi divided by three. , five pi divided by three. No solution

User Dhruvbird
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1 Answer

2 votes

Answer:


x=0,x=\pi,x=(\pi)/(3),x=(2\pi)/(3),x=(4\pi)/(3),x=(5\pi)/(3)

Explanation:

This is a trigonometric equation where we need to use the cosine of the double-angle formula


cos4x=2cos^22x-1

Replacing in the equation


cos4x - cos2x = 0

We have


2cos^22x-1 - cos 2x = 0

Rearranging


2cos^22x - cos 2x-1 = 0

A second-degree equation in cos2x. The solutions are:


cos2x=1,cos2x=-(1)/(2)

For the first solution

cos2x=1 we find two solutions (so x belongs to [0,2\pi))


2x=0, 2x=2\pi

Which give us


x=0,x=\pi

For the second solution


cos2x=-(1)/(2)

We find four more solutions


2x=(2\pi)/(3),2x=(4\pi)/(3),2x=(8\pi)/(3),2x=(10\pi)/(3)

Which give us


x=(\pi)/(3),x=(2\pi)/(3),x=(4\pi)/(3),x=(5\pi)/(3)

All the solutions lie in the interval
[0,2\pi)

Summarizing. The six solutions are


x=0,x=\pi,x=(\pi)/(3),x=(2\pi)/(3),x=(4\pi)/(3),x=(5\pi)/(3)

User Pradi KL
by
7.7k points

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