cos^2θ - sin^2θ = sinθ +1

Question

What is the solution set?

Answer 1

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.