Question

Find the values of ∈[0,] such that 3sin²−2cos³=0 .

Answer 1

To find the values of θ that satisfy the equation 3sin²θ - 2cos³θ = 0, we factor the equation and solve for θ, which gives us the solutions π/2 and 3π/2.

Step-by-step explanation:

The student asked to find the values of θ within the interval [0, 2π] that satisfy the equation 3sin²θ − 2cos³θ = 0. We can approach this by factoring and using trigonometric identities.

First, recognize that a factor of cos²θ can be taken out from the equation:

• 2cos³θ = 2cos²θ × cosθ
• 3sin²θ = 3(1 - cos²θ) since sin²θ = 1 - cos²θ

Now we rewrite the equation:

3(1 - cos²θ) - 2cos³θ = 0

Let's factor and solve for θ:

1. Multiply out and rearrange terms: 3 - 3cos²θ - 2cos³θ = 0
2. Factor out cos²θ: cos²θ(2cosθ + 3) - 3 = 0
3. Set each factor equal to zero and solve for θ: cos²θ = 0 or 2cosθ + 3 = 0
4. For cos²θ = 0: θ = π/2, 3π/2
5. For 2cosθ + 3 = 0: θ = cos⁻¹(-3/2), but since -3/2 is not within the range of the cosine function, there are no solutions from this part.

Hence, the values of θ that satisfy the equation are π/2 and 3π/2.