Question

4 cos²x – 1=0 What is the solution?

Answer 1

Answer:

$x=\frac{\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{5\pi }{3}+2\pi n, n \in \mathbb{Z}$

$x=\frac{2\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{4\pi }{3}+2\pi n, n \in \mathbb{Z}$

or

$x=\frac{\pi}{3}+\pi n, n \in \mathbb{Z}$

$x=\frac{2\pi }{3}+\pi n, n\in \mathbb{Z}$

Step-by-step explanation:

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

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

$cos(x)=\pm\frac{1}{2}$

So, when cos(x) is equal to

$\frac{1}{2} \text{ and } -\frac{1}{2}$    ?

For

$cos(x)=\frac{1}{2}$

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

$x=\frac{\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{5\pi }{3}+2\pi n, n \in \mathbb{Z}$

or

$x=\frac{\pi}{3}+\pi n, n \in \mathbb{Z}$

For

$cos(x)=-\frac{1}{2}$

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

$x=\frac{2\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{4\pi }{3}+2\pi n, n \in \mathbb{Z}$

or

$x=\frac{2\pi }{3}+\pi n, n\in \mathbb{Z}$