The given equation is in the form of a quadratic equation. We can rewrite it as:

4sin^2(θ) - 6sin(θ) - 4 = 0

Let's solve this equation step by step:

Step 1: Divide the entire equation by 4 to simplify it:

sin^2(θ) - 1.5sin(θ) - 1 = 0

Step 2: Now, let's use the quadratic formula to solve for sin(θ). The quadratic formula is given by:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

Here, a = 1, b = -1.5, and c = -1. Substituting these values into the quadratic formula gives:

sin(θ) = [1.5 ± sqrt((-1.5)^2 - 4*1*-1)] / (2*1)
= [1.5 ± sqrt(2.25 + 4)] / 2
= [1.5 ± sqrt(6.25)] / 2
= [1.5 ± 2.5] / 2

This gives us two possible solutions:

sin(θ) = 2 or sin(θ) = -0.5

Step 3: Now, we need to find the values of θ that satisfy these equations. Remember that the range of θ is 0 ≤ θ

Question

The given equation is in the form of a quadratic equation. We can rewrite it as:

4sin^2(θ) - 6sin(θ) - 4 = 0

Let's solve this equation step by step:

Step 1: Divide the entire equation by 4 to simplify it:

sin^2(θ) - 1.5sin(θ) - 1 = 0

Step 2: Now, let's use the quadratic formula to solve for sin(θ). The quadratic formula is given by:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

Here, a = 1, b = -1.5, and c = -1. Substituting these values into the quadratic formula gives:

sin(θ) = [1.5 ± sqrt((-1.5)^2 - 4*1*-1)] / (2*1)
       = [1.5 ± sqrt(2.25 + 4)] / 2
       = [1.5 ± sqrt(6.25)] / 2
       = [1.5 ± 2.5] / 2

This gives us two possible solutions:

sin(θ) = 2 or sin(θ) = -0.5

Step 3: Now, we need to find the values of θ that satisfy these equations. Remember that the range of θ is 0 ≤ θ < 2π.

For sin(θ) = 2, there are no solutions because the range of the sine function is -1 ≤ sin(θ) ≤ 1.

For sin(θ) = -0.5, the solutions are θ = 7π/6, 11π/6.

So, the solutions to the given equation are θ = 7π/6, 11π/6.

Knowee AI · Accepted Answer