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Solve 5sec(4x)=10

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Solve 5sec(4x)=10 ...[0,2pi)

asked Oct 30, 2014 in TRIGONOMETRY by anonymous

1 Answer

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The trigonometric equation is 5 sec(4x) = 10 and interval is [0, 2π).

Let 4x = t ⇒ 5 sec(t) = 10 ⇒ sec(t) = 2 ⇒ cos(t) = 1/2 ⇒ cos(t) = cos(π/3).

General solution: θ = 2nπ ± α, where n is an integer.

If α = π/3 ⇒ t = 2nπ ± π/3, where n is an integer.

4x = 2nπ ± π/3       [Since 4x = t]

x = nπ/2 ± π/12

If n = 0 ⇒ x = π/12

If n = 1 ⇒ x = π/2 - π/12 = 5π/12 and x = π/2 + π/12 = 7π/12.

If n = 2 ⇒ x = 2π/2 - π/12 = 11π/12 and x = 2π/2 + π/12 = 13π/12.

If n = 3 ⇒ x = 3π/2 - π/12 = 17π/12 and x = 3π/2 + π/12 = 19π/12.

If n = 4 ⇒ x = 4π/2 - π/12 = 23π/12.

The solutions in the interval [0, 2π) are x = π/12, x = 5π/12, 7π/12, 11π/12, 13π/12, 17π/12, 19π/12, and 23π/12.
answered Oct 30, 2014 by casacop Expert

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