solve for x between 0 and 2 pi :

Question

 2cos^2 x=1-cosx

Answer 1

Given that 2cos^2x = 1-cosx

2cos^2x + cosx - 1 = 0

2cos^2x + cosx - cosx -1 = 0

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

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

2cosx = 1           

cosx = 1/2

x = cos^-1( 1/2)

x = 60 ⁰

x =  π/3

cosx+1 = 0

cosx = -1

x = cos^-1( -1)

x = 180⁰

x = π

Therefore the values of  x = π/3 and π.

 

 

 

Comments

General solution : If cos(θ) = cos(∝) then θ = 2nπ ± ∝, where n is integer.

cos(x) = 1/2 and cos(x) = - 1

cos(x) = cos(π/3) and cos(x) = cos(π)

If ∝ = π/3 then x = 2nπ + π/3 and x = 2nπ - π/3.

If n = 0 then x = 2(0)nπ + π/3 = π/3

If n = 1 then x = 2(1)nπ - π/3 = 5π/3.

The solutions in the interval (0, 2π] are π/3, π and 5π/3.