a) To solve the equation -√12 csc (x- π/3) + 7 = 3, we need to isolate the variable x. Here's how you can solve it:
Step 1: Subtract 7 from both sides of the equation to get -√12 csc (x- π/3) = -4.
Step 2: Divide both sides of the equation by -√12 to get csc (x- π/3) = -4 / -√12, which simplifies to csc (x- π/3) = 4 / √12.
Now, to find the solutions on the given interval [-2π, 2π], we need to determine the values of (x-π/3) that satisfy this equation.
Step 3: Take the reciprocal of both sides to get sin (x- π/3) = √12 / 4, which simplifies to sin (x- π/3) = √3 / 2.
Step 4: Solve for x- π/3 by taking the inverse sin (sin^(-1)) of both sides, giving us x- π/3 = sin^(-1)(√3 / 2).
Step 5: Now, we need to find the general solutions for x within the given interval. We can write those solutions as x = π/3 + nπ + sin^(-1)(√3 / 2), where n is an integer.
To get the approximate solutions, we can use a calculator to evaluate sin^(-1)(√3 / 2) and find the corresponding values for x within the interval [-2π, 2π].
b) To solve the equation 2sin^2 θ = 1 - sin θ, we need to isolate the variable θ. Here's how you can solve it:
Step 1: Rearrange the equation to get 2sin^2 θ + sin θ - 1 = 0.
Step 2: Factor the quadratic equation. In this case, we need to find two numbers whose product is -2 and whose sum is 1. The factors are 2 and -1. Therefore, we can rewrite the equation as (2sin θ + 1)(sin θ - 1) = 0.
Now, we have two possible equations that can provide solutions for θ: 2sin θ + 1 = 0 or sin θ - 1 = 0.
Step 3: Solve the first equation, 2sin θ + 1 = 0.
- Subtract 1 from both sides: 2sin θ = -1.
- Divide both sides by 2: sin θ = -1/2.
Step 4: Solve the second equation, sin θ - 1 = 0.
- Add 1 to both sides: sin θ = 1.
Now, we need to find the solutions for θ within the given interval [-3π/2, 2π]. We can use a calculator to determine the corresponding values of θ for sin θ = -1/2 and sin θ = 1 within the interval.
Note: When solving trigonometric equations, it's always a good idea to check for extraneous solutions by substituting the values back into the original equation to ensure they satisfy the equation.