Solve cos2x=-2?v/2 such that 0=x=2p.

Question

i'm not sure hot to get to the answer

Answer 1

cos(2x) = -√2/2

 

2x = 3π/4 + 2kπ or 5π/4 + 2kπ, k = 0, ±1, ±2, ...

 

x = 3π/8 + kπ or 5π/8 + kπ

 

If k = 0, we get x = 3π/8, 5π/8

If k = 1, we get x = 11π/8, 13π/8

 

Other choices for k give solutions outside of the interval [0, 2π].

 

Solutions:  3π/8, 5π/8, 11π/8, 13π/8

 

Answer 2

cos 2x = -2

 

1 - 2sin²x = - 2

 

 

                   3 = 2sin²x

 

                3/2 = sin²x

 

 (+ or - √(3/2) = sinx

 

√(3/2) = 1.22

 

Since: -1 ≤ sin x ≤ 1  there is no solution to the problem

as I wrote it down, once again I did not know what ?v/2 meant in the equation.