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Find all solutions to 2sin(θ)=√3 on the interval 0≤θ<2π

Sagot :

Answer:

2sin(3θ) - √3 = 0

Step-by-step explanation:

 sin(3θ) = √3/2

 

3θ = π/3 + 2kπ or 2π/3 + 2kπ, k = 0, ±1, ±2, ±3,...

 

θ = π/9 + 2kπ/3, 2π/9 + 2kπ/3

 

If k = 0, we get θ = π/9, 2π/9

If k = 1, we get θ = 7π/9, 8π/9

If k = 2, we get θ = 13π/9, 14π/9

 

Other values of k give values of θ lying outside of the interval [0, 2π).

Facundo3141592idn

We want to find all the solutions of 2sin(θ) = √3 on the interval 0 ≤ θ ≤ 2π. The only solution is θ = 1.05

So we want to solve the equation: 2sin(θ) = √3

First, we divide both sides by 2 to get:

sin(θ) = (√3)/2

Now remember the inverse sine function, it acts as follows:

Asin(sin(x)) = xsin(Asin(x)) = x

So we can apply this to both sides to get:

Asin(sin(θ)) = Asin((√3)/2)θ = Asin((√3)/2) = 1.05

And because we know that the period of the sine function is 2π, we know that there is only one solution on the range between 0 and 2π, then the only solution in the interval is:

θ = 1.05

If you want to learn more, you can read:

https://brainly.com/question/12015707

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