I am trying to solve \sqrt{3}\tan\theta=2\sin\theta on the interval [-\pi,\pi].

haugmbd

haugmbd

Answered question

2022-02-28

I am trying to solve 3tanθ=2sinθ on the interval [π,π].

Answer & Explanation

aceptanteppt

aceptanteppt

Beginner2022-03-01Added 5 answers

3tanθ=2sinθ
3sinθcosθ=2sinθ
3sinθ2sinθcosθ=0
sinθ(32cosθ)=0
sinθ=0, cosθ=32
θ=kπ, θ=2kπ±π6
For given interval, θ[π,π], we get
θ=π, π6, 0, π6, π

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