How do I solve these equations:

Mathematics

1 answer:

igor_vitrenko [27]3 years ago

7 0

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°

_______________________________________________________

$sin 2\theta=\sqrt{3cos\theta}$

By squaring both sides

$sin^2 2\theta={3cos\theta}$

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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