Over the interval [0, 2π), what are the solutions to cos(2x) = cos(x)? Check all that apply. 0

StartFraction pi Over 3 EndFraction

StartFraction 2 pi Over 3 EndFraction

Pi

StartFraction 4 pi Over 3 EndFraction

StartFraction 5 pi Over 3 EndFraction

Step-by-step explanation:

given the expression;

cos(2x) = cos(x)

According to trig identity;

cos(2x) = cos(x+x)

cos(2x) = cos x cos x - sinx sinx

cos(2x) = cos²(x)-sin²(x)

cos(2x) = cos²(x)-(1-cos²x)

cos(2x) = cos²(x)+cos²x-1

cos(2x) = 2cos²(x)-1

2cos²(x)-1 = cos(x)

let P = cosx

2P²-1 = P

2P²-P-1 = 0

Factorize;

2P²-2P+P-1 = 0

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

2P - 1 = 0 and P-1 = 0

P = 1/2 and 1

cosx = 1/2 and cos x = 1

x = arccos 1/2

x = π/3

Also;

x = arccos1

x = 0

Hence the value of x are 0 and  π/3

Also the angle = π+ π/3 = 4π/3

The angles are 0, π/3 and 4π/3

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake