Question

sendo U=(0, 2pi) calcule: cos2x-sen2x=0​

Answer 1

Resposta:

S = {π/8, 5π/8, 9π/8, 13π/8}

Explicação passo-a-passo:

Sabemos que se senα = senβ ⇒ α = β  + 2kπ ou x = π - β + 2kπ

Sabemos também que cosα = sen(π/2 - α)

cos2x - sen2x = 0

cos2x = sen2x

sen2x = cos2x

sen2x = sen(π/2 - 2x)

2x = π/2 - 2x + 2kπ

2x + 2x = π/2 + 2kπ

4x = π/2 + 2kπ  (dividir por 4)

x = π/8 + kπ/2

ou

2x = π - (π/2 - 2x) + 2kπ

2x = π - π/2 + 2x + 2kπ

2x - 2x = π/2 + 2kπ

0x = π/2 + 2kπ

Não existe x

x = π/8 + kπ/2 , com k ∈ Z

p/k = 0 ⇒ x = π/8 + 0.π/2 ⇒ x = π/8

p/k = 1 ⇒ x = π/8 + 1.π/2 ⇒ x = π/8 + π/2 ⇒ x = π/8 + 4π/8 ⇒ x = 5π/8

p/k = 2 ⇒ x = π/8 + 2.π/2 ⇒ x = π/8 + π ⇒ x = π/8 + 8π/8 ⇒ x = 9π/8

p/k = 3 ⇒ x = π/8 + 3.π/2 ⇒ x = π/8 + 12π/8 ⇒ x = 13π/8

p/k = 4 ⇒ x = π/8 + 4.π/2 ⇒ x = π/8 + 16π/2 ⇒ x = π/8 + 2π (não serve, pois ultrapassa o intervalo [0, 2π]

S = {π/8, 5π/8, 9π/8, 13π/8}