# Find all solutions in the interval [0, 2π], cos^{2}x + 2 cosx + 1 = 0

**Solution:**

Given Equation is:

cos^{2}x + 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], cos^{2}x + 2 cosx + 1 = 0 is x = π.

## Find all solutions in the interval [0, 2π], cos^{2}x + 2 cosx + 1 = 0

**Summary:**

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

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