Question

Use a Double- or Half-Angle Formula to solve the equation in the interval [0,2π ). (Enter your answers as a comma-separated list.)

sin(3θ)−sin(6θ)=0
θ=
3
2πk
​
,
9
π
​
+
3
2πk
​
,
3
π
​
+
3
2πk
​
,
9
5π
​
+
3
2πk
​

​

Answer 1

θ = 0, π/9, π/3, 2π/3, 5π/9, 2π

To solve sin(3θ)−sin(6θ)=0, apply the double-angle formula, yielding sin(3θ)(1 - 2cos(3θ)) = 0. Set each factor to zero to find the solutions within [0,2π).

Step-by-step explanation:

To solve the equation sin(3θ)−sin(6θ)=0 in the interval [0,2π), we can use the double-angle formula for sine, which states that sin(2α) = 2sin(α)cos(α). Applying this to sin(6θ), we get sin(6θ) = 2sin(3θ)cos(3θ).

Now, substitute this expression back into the original equation: sin(3θ) - 2sin(3θ)cos(3θ) = 0. Factor out sin(3θ) to get sin(3θ)(1 - 2cos(3θ)) = 0.

Now, set each factor equal to zero:

1. sin(3θ) = 0, which gives us solutions θ = 0, π, and 2π.

2. 1 - 2cos(3θ) = 0. Solve for cos(3θ), and you'll find cos(3θ) = 1/2, which occurs when 3θ = π/3, 5π/3. This gives additional solutions θ = π/9, 5π/9.

Combining all solutions within the interval [0,2π), we get θ = 0, π/9, π/3, 2π/3, 5π/9, and 2π. These values satisfy the given equation sin(3θ)−sin(6θ)=0 within the specified interval.

In conclusion, the solutions are θ = 0, π/9, π/3, 2π/3, 5π/9, and 2π.

Answer 2

To solve sin(3θ) - sin(6θ) = 0 using a double or half-angle formula, we can use the double-angle formula for sine and factor the equation to find the solutions for θ. The solutions are θ = πk, θ = 3π/6 + 2πk, θ = 5π/6 + 2πk, and θ = 9π/6 + 2πk, where k is an integer.

Step-by-step explanation:

To solve the equation sin(3θ) - sin(6θ) = 0 using a double or half-angle formula, we can start by using the double-angle formula for sine: sin(2θ) = 2sin(θ)cos(θ). Applying this formula, we have sin(6θ) = 2sin(3θ)cos(3θ). Substituting this into the original equation, we get:

sin(3θ) - 2sin(3θ)cos(3θ) = 0

Factoring out sin(3θ), we have sin(3θ)(1 - 2cos(3θ)) = 0.

So, either sin(3θ) = 0 or 1 - 2cos(3θ) = 0.

If sin(3θ) = 0, θ must be πk, where k is an integer.

If 1 - 2cos(3θ) = 0, then cos(3θ) = 1/2. Using the inverse cosine function, we find that θ = 3π/6 + 2πk, θ = 5π/6 + 2πk, and θ = 9π/6 + 2πk.

Therefore, the solutions for θ in the interval [0, 2π) are: θ = πk, θ = 3π/6 + 2πk, θ = 5π/6 + 2πk, and θ = 9π/6 + 2πk, where k is an integer.