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cos^2θ - sin^2θ = sinθ +1

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What is the solution set?

asked Dec 8, 2014 in TRIGONOMETRY by anonymous

1 Answer

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The trigonometric equation is cos2 θ - sin2 θ = sin θ + 1.

Pythagorean identity : sin2 θ + cos2 θ = 1.

1 - sin2 θ - sin2 θ = sin θ + 1

1 - 2sin2 θ = sin θ + 1

2sin2 θ + sin θ = 0
sin θ(2sin θ + 1) = 0

Apply zero product property.

sin θ = 0 and 2sin θ + 1 = 0.

  • sin θ = 0.

sin(θ) = sin(0)

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

θ = nπ + (- 1)n(0)

θ = nπ.

  • 2sin (θ) + 1 = 0.

2sin (θ) = - 1

sin (θ) = - 1/2

sin(θ) = sin(- π/6)

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

θ = nπ + (- 1)n(- π/6).

The solution set is {θ / θ = nπ and θ = nπ + (- 1)n(- π/6)}, where n is an integer.

answered Dec 8, 2014 by lilly Expert

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