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