First, let's solve this quadratic equation for y = cos(2x):

`2y^2 +y - 1 = 0`

This equation can be solved by factoring:

(2y - 1)(y + 1 ) = 0

2y = 1 and y = -1

These two values of y give us two trigonometric equations:

2cos(2x) =...

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First, let's solve this quadratic equation for y = cos(2x):

`2y^2 +y - 1 = 0`

This equation can be solved by factoring:

(2y - 1)(y + 1 ) = 0

2y = 1 and y = -1

These two values of y give us two trigonometric equations:

2cos(2x) = 1 and cos(2x) = -1

Consider the first equation first. Divide its both sides by 2:

`cos(2x) = 1/2`

Cosine of 2x is 1/2 when `2x = +-pi/3 + 2pik`

To find the possible values of x, divide by 2:

`x = +-pi/6 + pik`

Which of these values of x will lie within the interval `(0, 2pi)?`

`pi/6, (5pi)/6, (7pi)/6, (11pi)/6`

The reason you are told that the interval will "switch" to `(0, 4pi)`

is because if x belongs to `(0, 2pi),`

then 2x belongs to `(0, 4pi)`

and we could have limited the values of 2x based on that. But it is easier just to solve for all possible value of x and then restrict them to those that lie within `(0, 2pi).`

The second equation was cos(2x) = -1. This happens when

`2x = pi + 2pik`

Divide by 2: `x = pi/2 + pik`

The values of x between 0 and `2pi`

are `pi/2, (3pi)/2`

**So these are solutions of the given quadratic trigonometric equation that are within `(0, 2pi):` **

**`pi/2, (3pi)/2, pi/6, (5pi)/6, (7pi)/6, (11pi)/6` **