Solve each equation on the given interval. If answers can be exact, leave the exact. If approximations are required, round solutions to the nearest thousandth.
a) -√12 csc (x- pi/3) +7 = 3' XE [-2 pi, 2 pi]
b) 2sin^2 theta = 1 - sin theta, X E [-3pi/2, 2pi]
2sin^2 θ = 1 - sin θ
This one is easier ...
2sin^2 θ + sin θ - 1 = 0
(2sinθ - 1)(sinθ + 1) = 0
sinθ = 1/2 or sinθ = -1
I will work in degrees for this one, then switch to radians
θ = 30° or θ = 150°
or
θ = 270°
but we have a domains -360° ≤ θ < 360°
so θ = -330°, -210, -90°, 30°, 150°, 270°
in radians:
θ = -11π/6, -7π/6, -π/2, π/6, 5π/6, 3π/2
-√12 csc (x- pi/3) +7 = 3
-√12 csc (x- pi/3) = -4
csc (x- pi/3) = -4/2√3 = 2/√3
sin (x- pi/3) = √3/2 , so x-π/3 must be in quad I or II
I know sin π/3 = √3/2
then (x-π/3) = π/3 or x-π/3 = π - π/3
x = π/3 + π/3
x = 2π/3
or x-π/3 = π - π/3
x = 2π/3 + π/3 = π
now the period of sin (x- pi/3) is 2π, so another answer would be
2π/3 - 2π = -4π/3
π - 2π = -π
so for the given domain:
x = -4π/3, -π, 2π/3, and π
Thank you!!
I really couldn't solve this one sorry everyone
we appreciate you mathhelper
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.