Respuesta :

MathPhys

Answer:

0, π/6, 5π/6, π

Step-by-step explanation:

cos² θ − sin² θ = 1 − sin θ, 0 ≤ θ < 2π

Use Pythagorean identity:

(1 − sin² θ) − sin² θ = 1 − sin θ

1 − 2sin² θ = 1 − sin θ

2sin² θ = sin θ

2sin² θ − sin θ = 0

sin θ (2 sin θ − 1) = 0

sin θ = 0

θ = 0 or π

2 sin θ − 1 = 0

sin θ = 1/2

θ = π/6 or 5π/6

sqdancefan

Answer:

  θ ∈ {0, π/6, 5π/6, π, 2π}

Step-by-step explanation:

Use the equivalent for cos² and solve the resulting quadratic in sin.

  (1 -sin(θ)²) -sin(θ)² = 1 -sin(θ)

  0 = 2sin(θ)² -sin(θ) . . . . . subtract the left side

  0 = sin(θ)(2sin(θ) -1)

This has solutions ...

  sin(θ) = 0   ⇒   θ = {0, π, 2π}

  sin(θ) = 1/2   ⇒   θ = {π/6, 5π/6}

The solution set is ...

  θ ∈ {0, π/6, 5π/6, π, 2π}

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