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How do I solve for x on the interval [0, 2pi]

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4cos^2 x - 3 = 0?

asked Aug 9, 2014 in TRIGONOMETRY by anonymous

1 Answer

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The trigonometric equation is 4cos2 x - 3 = 0.

Add 3 to each side.

4cos2 x = 3

Divide each side by 4.

cos2 x = 3/4

cos x = ±√3/2.

⇒ cos x = √3/2  and  cos x = - √3/2.

  • cos (x) = √3/2.

cos (x) = cos(π/6)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒ x = 2nπ ± π/6

If n = 0, x = 2(0)π + (π/6) and x = 2(0)π - (π/6) = π/6 and - π/6.

If n = 1, x = 2(1)π + (π/6) and x = 2(1)π - (π/6) = 2π + π/6 and 2π - π/6 = 13π/6 and 11π/6,

Therefore, the solutions of the given equation are x = π/6 and x = 11π/6 in the interval [0, 2π].

  • cos (x) = - √3/2.

cos (x) = cos(5π/6)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒ x = 2nπ ± 5π/6

If n = 0, x = 2(0)π + (5π/6) and x = 2(0)π - (5π/6) = 5π/6 and - 5π/6.

If n = 1, x = 2(1)π + (5π/6) and x = 2(1)π - (5π/6) = 2π + 5π/6 and 2π - 5π/6 = 17π/6 and 7π/6,

Therefore, the solutions of the given equation are x = 5π/6 and x = 7π/6 in the interval [0, 2π].

The solutions of the given equation are x = π/6, x = 5π/6, x = π, x = 7π/6 and x = 11π/6 in the interval [0, 2π].

answered Aug 9, 2014 by lilly Expert

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