ramose7368 ramose7368
  • 10-08-2020
  • Mathematics
contestada

How do I solve these equations:
sin(2θ) + sin θ = 0
sin(2θ) = sqrt 3 cos θ

for intervals 0=< θ < 2pi

Respuesta :

StaceeLichtenstein StaceeLichtenstein
  • 11-08-2020

Answer:

Step-by-step explanation:

Given that

sin(2θ)+sinθ=0

We know that

sin(2θ)=2 sinθ x cosθ

Therefore

2 sinθ x cosθ + sinθ=0

sinθ(2 cosθ+1)=0

sinθ= 0

θ=0

2 cosθ+1=0

cosθ= - 1/2

θ=120°

_______________________________________________________

[tex]sin 2\theta=\sqrt{3cos\theta}[/tex]

By squaring both sides

[tex]sin^2 2\theta={3cos\theta}[/tex]

4 sin²θ x cos²θ=3 cosθ

4 sin²θ x cos²θ - 3 cosθ=0

cos θ = 0

θ= 90°

4 sin²θ=3

θ=60°

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