Find all solutions in the interval [0, 2π], cos²x + 2 cosx + 1 = 0

Solution:

Given Equation is:

 cos²x + 2 cosx + 1 = 0

We can write the above equation using algebraic identity as:

⇒ (cos x + 1)² = 0

⇒ cos x + 1 = 0

⇒ cos x = -1

Therefore,

cos x = cos π

General solution is x = 2nπ ± π where n ∈ Z.

Now put n = 0, x = ±π

⇒ x = π and x = -π

But x = -π ∉ [0, 2π]

n = 1; x = 2(1)π ± π 

• x = 2π + π = 3π ∉ [0, 2π]
• x = 2π - π = π ∈ [0, 2π]

Therefore, solution in the interval [0, 2π], cos²x + 2 cosx + 1 = 0 is x = π.

Find all solutions in the interval [0, 2π], cos²x + 2 cosx + 1 = 0

Summary:

The solution in the interval [0, 2π], cos²x + 2 cosx + 1 is x = π.  General solution is x = 2nπ ± π where n ∈ Z.