Question

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

Answer 1

Answer:

$x=0,x=\pi,x=\frac{\pi}{3},x=\frac{2\pi}{3},x=\frac{4\pi}{3},x=\frac{5\pi}{3}$

Step-by-step 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:

$cos2x=1,cos2x=-\frac{1}{2}$

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

$cos2x=-\frac{1}{2}$

We find four more solutions

$2x=\frac{2\pi}{3},2x=\frac{4\pi}{3},2x=\frac{8\pi}{3},2x=\frac{10\pi}{3}$

Which give us

$x=\frac{\pi}{3},x=\frac{2\pi}{3},x=\frac{4\pi}{3},x=\frac{5\pi}{3}$

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

Summarizing. The six solutions are

$x=0,x=\pi,x=\frac{\pi}{3},x=\frac{2\pi}{3},x=\frac{4\pi}{3},x=\frac{5\pi}{3}$