An equation is given. (Enter your answers as a comma-separated list. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)

2 cos 2θ − 1 = 0

a). Find all solutions of the equations

answer--> θ= (π+6nπ)/6 , (5π+6nπ)/6

b). Find the solutions in then interval [0,2π).

θ=___________

To find all solutions of the equation 2 cos 2θ − 1 = 0, we can solve for θ by isolating cos 2θ:

2 cos 2θ = 1

Dividing both sides by 2 gives us:

cos 2θ = 1/2

a) Now, we need to find all solutions for θ. We can use the inverse cosine function (also known as arccos) to find the solutions.

The arccosine of 1/2 is π/3 (or 60 degrees). Since the cosine function has a period of 2π, we can find all the solutions by adding integer multiples of the period to π/3.

So, the solutions for θ are:

θ = (π + 6nπ)/6, where n is an integer.

b) To find the solutions in the interval [0, 2π), we need to restrict the values of n to produce solutions that fall within this interval.

By plugging in the values of n, we get:

θ1 = (π + 6(0)π)/6 = π/6
θ2 = (π + 6(1)π)/6 = 7π/6
θ3 = (π + 6(2)π)/6 = 13π/6

Among these solutions, only θ1 = π/6 falls within the interval [0, 2π).

Thus, the solution in the interval [0, 2π) is:

θ = π/6.

To find the solutions to the equation 2 cos 2θ - 1 = 0, we'll follow these steps:

Step 1: Solve for cos 2θ
Add 1 to both sides of the equation:
2 cos 2θ = 1

Step 2: Divide both sides by 2
cos 2θ = 1/2

Step 3: Find the inverse cosine of both sides
Using the inverse cosine function, which is usually written as cos^(-1) or arccos, we have:
2θ = cos^(-1)(1/2)

Step 4: Determine the general solutions for θ
To find the general solutions for θ, we need to consider the values of θ that satisfy the equation.

First, let's find the reference angle, which is the angle between 0 and π/2 that has the same cosine value as our equation. So, cos^(-1)(1/2) equals π / 3.

Now, we can find the general solutions for θ by considering the periodic nature of cosine.

Since cos is positive in the first and fourth quadrants, we have two sets of solutions:

Set 1: 2θ = π/3 + 2πn
To find θ, divide both sides by 2:
θ = (π/3 + 2πn) / 2
Simplifying the expression, we get θ = (π + 6πn) / 6.

Set 2: 2θ = -π/3 + 2πn
Divide both sides by 2:
θ = (-π/3 + 2πn) / 2
Simplifying, we get θ = (5π + 6πn) / 6.

So, the general solutions for θ are θ = (π+6πn)/6 and θ = (5π+6πn)/6, where n is an integer.

b). Find the solutions in the interval [0, 2π).
To find the solutions within the specific interval [0, 2π), we need to find the values of θ that fall within this range.

Substitute n = 0 into θ = (π+6πn)/6:
θ = (π+6π*0)/6 = π/6

Substitute n = 1 into θ = (π+6πn)/6:
θ = (π+6π*1)/6 = (7π)/6

Since π/6 and 7π/6 both fall within the interval [0, 2π), the solutions are:
θ = π/6, 7π/6.

Looks ok to me. It might be easier to read

π/6 + nπ
5π/6 + nπ

So, plug in various values of n to get all the solutions in [0,2π)

There will be four of them