Find all solutions to the equation in the interval [0, 2π). cos 4x - cos 2x = 0

See below in bold.

Step-by-step explanation:

cos 4x = 2cos^2 2x - 1 =  2 (2 cos^2 x - 1)^2 - 1

and cos 2x = 2 cos^2 x - 1 so we have:

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

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

(2 cos^2 x - 1)^2 -  cos^2 x = 0

Let c = cos^2 x, then:

(2c - 1)^2 - c = 0

4c^2 - 4c + 1 - c = 0

4c^2 - 5c + 1 = 0

c = 0.25, 1

cos^2 x  = 0.25 gives cos x = +/- 0.5

and cos^2 x = 1 gives cos x = +/- 1.

So for  x = +/- 1 , x = 0, π.

For cos x = +/- 0.5, x = π/3, 2π/3, 4π3,5π/3.

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5

addition

step-by-step explanation: