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How do I find all solution of the equation

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sin^2 (3x) = 1 ?

asked Aug 9, 2014 in TRIGONOMETRY by anonymous

1 Answer

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The trigonometric equation is sin2 (3x) = 1.

⇒ sin (3x) = ± 1

sin (3x) = - 1 and sin (3x) = + 1.

  • sin (3x) = - 1.

sin(3x) = sin(3π/2)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)nα, where n is an integer.

 3x = nπ + (- 1)n(3π/2)

⇒ x = [nπ + (- 1)n(3π/2)]/3.

  • sin (3x) = 1.

sin(3x) = sin(π/2)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)nα, where n is an integer.

3x = nπ + (- 1)n(π/2)

⇒ x = [nπ + (- 1)n(π/2)]/3.

The solutions of the given equation are x = [nπ + (- 1)n(3π/2)]/3 and x = [nπ + (- 1)n(π/2)]/3, where n is an integer.

answered Aug 9, 2014 by lilly Expert

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