Question

Solve for θ if −6sinθ+4=3√3 +4 and 0≤θ<2π.

Answer 1

To solve for θ, subtract 4 from both sides to get −6sinθ = 3√3. Divide both sides by -6 to obtain sinθ = -1√2/2. Use the inverse sine function to find the value of θ. θ = sin^-1(-1√2/2). Simplify the value of θ.

Step-by-step explanation:

To solve for θ, we will isolate the variable by performing algebraic operations.

1. Subtract 4 from both sides to get −6sinθ = 3√3.

2. Divide both sides by -6 to obtain sinθ = -1√2/2.

3. Use the inverse sine function to find the value of θ. θ = sin^-1(-1√2/2).

4. Simplify the value of θ. θ = -π/4

Answer 2

To solve for θ in the equation -6sinθ+4=3
`√(3)` +4, isolate the term with sinθ, divide by -6 to solve for sinθ, and find the principal angle within the given range of 0≤θ<2π.

Step-by-step explanation:

To solve for θ in the equation -6sinθ+4=3
`√(3)` +4, we first isolate the term with sinθ by subtracting 4 from both sides: -6sinθ = 3√3. Then, divide both sides by -6 to solve for sinθ: sinθ = (3
`√(3)`)/-6 = -
`√(3/2)`. We can use the inverse sine function to find the value of θ: θ =
`sin^{(-1)`(-
`√(3/2)`).

Since we are given that 0≤θ<2π, we need to find the principal angle whose sine is -
`√(3/2)` within this range. The principal angle in the second quadrant is π/3. Therefore, the solution for θ is θ = π/3.