Question

4 cos²x – 1=0

What is the solution?

Answer 1

`$x=(\pi )/(3)+2\pi n, n\in \mathbb{Z}$`

`$\:x=(5\pi )/(3)+2\pi n, n \in \mathbb{Z}$`

`$x=(2\pi )/(3)+2\pi n, n\in \mathbb{Z}$`

`$\:x=(4\pi )/(3)+2\pi n, n \in \mathbb{Z}$`

or

`$x=(\pi)/(3)+\pi n, n \in \mathbb{Z} $`

`$x=(2\pi )/(3)+\pi n, n\in \mathbb{Z}$`

Explanation:

`4\text{cos}^2(x)-1=0\\4\text{cos}^2(x)=1\\`

`$cos(x)=\pm\sqrt{(1)/(4) } $`

`$cos(x)=\pm(1)/(2) $`

So, when cos(x) is equal to

`$(1)/(2) \text{ and } -(1)/(2)$` ?

For

`$cos(x)=(1)/(2) $`

We are talking about x = 60º and x = 300º, Quadrant I and IV, respectively. In radians:

`$x=(\pi )/(3)+2\pi n, n\in \mathbb{Z}$`

`$\:x=(5\pi )/(3)+2\pi n, n \in \mathbb{Z}$`

or

`$x=(\pi)/(3)+\pi n, n \in \mathbb{Z} $`

For

`$cos(x)=-(1)/(2) $`

We are talking about x = 120º and x = 240º, Quadrant II and III, respectively. In radians:

`$x=(2\pi )/(3)+2\pi n, n\in \mathbb{Z}$`

`$\:x=(4\pi )/(3)+2\pi n, n \in \mathbb{Z}$`

or

`$x=(2\pi )/(3)+\pi n, n\in \mathbb{Z}$`