How do I solve the following?
sec 3 theta/2 = 2 on the interval 0<theta<2 pi?
I tried to look in my textbook but it doesn't have an example. I also tried google and I was unsuccessful> How do I do this question? I know that the period for sec is 2 pi but I don't know how to figure out the interval it is positive 2 on. Any help is appreciated. Thank you.
To solve the equation sec(3θ/2) = 2 on the interval 0 < θ < 2π, you can follow these steps:
Step 1: Rewrite the equation using the reciprocal identity for sec.
cosec(3θ/2) = 1/2
Step 2: Convert cosec to sin.
sin(3θ/2) = 1/2
Step 3: Determine the solutions for sin(3θ/2) = 1/2 in the interval 0 < θ < 2π.
To find the solutions, you can use the unit circle or reference angles to identify values of θ that satisfy the equation.
Since sin(π/6) = 1/2, one solution is 3θ/2 = π/6.
To find the other solutions, you can add multiples of the period (2π) to the angle:
3θ/2 = π/6 + 2πk
where k is an integer representing the number of periods added.
Step 4: Solve the equation for θ.
To isolate θ, you can multiply both sides of the equation by 2/3:
θ = (π/6 + 2πk)(2/3)
Simplifying the expression:
θ = π/9 + 4πk/3
This equation represents the general solution for θ in the given interval. To find specific values of θ within the interval 0 < θ < 2π, you can substitute different values for k and simplify the expression.
Remember to check if any values of θ yield undefined values for sec(3θ/2) and exclude them from the final solution.
To solve the equation sec(3θ/2) = 2 on the interval 0 < θ < 2π, you can follow these steps:
Step 1: Start by isolating sec(3θ/2) on one side of the equation. Divide both sides by 2:
sec(3θ/2) = 2/1
sec(3θ/2) = 2
Step 2: To find the interval where sec(3θ/2) = 2, you need to recall the definition of secant function. Secant is the reciprocal of cosine, so you can rewrite the equation as:
cos(3θ/2) = 1/2
Step 3: Next, you can solve for cos(3θ/2) = 1/2 using inverse cosine or the cosine function's periodicity and special angle properties.
Since the range of inverse cosine (also known as arccos) is 0 to π, you should look for angles between 0 and π that have a cosine value of 1/2.
Step 4: You can use the unit circle or reference angles to determine the potential solutions within the specified interval.
The cosine function has a value of 1/2 at two different angles in the 0 to π range: π/3 and 5π/3.
Step 5: Substitute these potential values back into the original equation to check if they are valid solutions. Plug in the values of π/3 and 5π/3 for 3θ/2 and check if cos(3θ/2) = 1/2 holds true.
For π/3:
cos(3(π/3)/2) = cos(π/2) = 0 ≠ 1/2
For 5π/3:
cos(3(5π/3)/2) = cos(15π/6) = cos(5π/2) = 0 ≠ 1/2
Step 6: Since the equation does not hold true for either of the potential solutions, sec(3θ/2) = 2 does not have solutions within the specified interval 0 < θ < 2π.
Therefore, the equation has no solutions in the given interval.
if sec 3Ø/2 = 2 then
cos 3Ø/2 = 1/2
I know cos π/3 = 1/2
so
3Ø/2 = π/3 or 3Ø/2 = 2π - π/3 = 5π/3
then
3Ø = 2π/3 or 3Ø = 10π/3
Ø = 2π/9 or Ø = 10π/9 , both within the interval given
The period of the original is 2π/(3/2) = 4π/3
adding this period to any of our answers will produce more answers,
so another value of Ø = 2π/9 + 4π/3 = 14π/9
so Ø = 2π/3, 12π/9, and 14π/9
(120, 240, and 280 degrees, all 3 work in the original equation and are between 0 and 360 degrees)