prove the solution set for: ??????????

Question

 2 cos^2 + cos x = 1

Answer 1

The given equation is 2 cos²x + cosx = 1

2 cos²x + cosx - 1 = 0

2 cos²x + 2cosx - cosx - 1 = 0

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

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

cosx = 1/2 , cosx = -1

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

x = 60 and  x = 180

Therefore x = π/3 and x = π.

Comments

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

cos(x) = cos(π/3) and cosx = cos(π)

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

If ∝ = π/3 then x = 2nπ ±  π/3.

If ∝ = π then x = 2nπ ±  π.

The general solutions are x = 2nπ ±  π/3 and x = 2nπ ±  π.

Answer 2

The given trignometric equation is 2cos² + cosx = 1

substract 1 from each side

2cos²x + cosx - 1 = 0

2cos²x  + 2cosx - cosx -1 = 0

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

Take out common factor

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

Apply zero product property

2cosx - 1 =0                                                cosx + 1 =0

cosx = 1/2                                                    cosx = -1

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

 x = 60⁰                                                            x =  180⁰        

x = 60⁰  (or) 180⁰