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A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
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.
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