interval [0,2pi) 2 cos 2x - (root2)=0

Question

Please help me find this answer

Answer 1

2 cos 2x - √2 = 0

 

Add √2 to both sides:

2 cos 2x = √2

 

Divide both sides by 2:

cos 2x = √2/2

 

In the interval [0,2π), cosine is √2/2 at π/4 and 7π/4.  But since you're taking the cos of 2x, you're going to end up dividing the angle by 2.  So you have to extend the interval where you look for cosine equal  to √2/2 out to 4π.  So you also have to look at 9π/4 and 15π/4.  (Those are π/4 and 7π/4 with 2π added to each.)

 

So the possible values of 2x are π/4, 7π/4, 9π/4 and 15π/4.

 

So x = π/8, 7π/8, 9π/8 or 15π/8.

 

 

 

Answer 2

2 cos 2x = √2

cos 2x = √2 / 2

cos^-1 √2 / 2 = π/4, 5π/4

2x = π/4, 5π/4

x = π/8, 5π/8