Find all solutions of 4 (cos (x)**2)-1=0 in the interval (6pi, 8pi).
(Leave your answers in exact form and enter them as a comma-separated list.)
4 cos^2 x - 1 = 0
cos^2 x = 1/4
cosx = ± 1/2
x = 60° , 120° , 240° , 300°
change to radians, I will let you decide how many periods (2π) have to be added to each answer for your domain
To find the solutions of the equation 4(cos(x)**2) - 1 = 0 in the interval (6π, 8π), we can follow these steps:
Step 1: Solve for cos(x) by isolating the cosine term.
4(cos(x)**2) - 1 = 0
Add 1 to both sides of the equation:
4(cos(x)**2) = 1
Divide both sides by 4:
cos(x)**2 = 1/4
Step 2: Take the square root of both sides to solve for cos(x).
cos(x) = ±√(1/4)
cos(x) = ±1/2
Step 3: Find the values of x in the given interval where cos(x) equals ±1/2.
For cos(x) = 1/2:
- In the interval (6π, 8π), there is one solution within this range:
x = 7π/3
For cos(x) = -1/2:
- In the interval (6π, 8π), there is one solution within this range:
x = 11π/6
So, the solutions in the interval (6π, 8π) are:
x = 7π/3, 11π/6
To find all the solutions of the equation 4(cos(x)^2) - 1 = 0 in the interval (6pi, 8pi), we can start by isolating the cosine term and then solve for x using the inverse cosine function.
First, let's add 1 to both sides of the equation:
4(cos(x)^2) = 1
Next, divide both sides of the equation by 4:
cos(x)^2 = 1/4
To solve for cos(x), we can take the square root of both sides:
cos(x) = ±√(1/4)
Since cos(x) can be both positive and negative, we need to consider both cases.
Case 1: cos(x) = √(1/4)
To find the value of x, we can take the inverse cosine of both sides:
x = arccos(√(1/4))
Using a calculator, we find that the principal value of arccos(√(1/4)) is Pi/3. However, we need to find the solutions in the interval (6pi, 8pi).
Since the cosine function is periodic with a period of 2pi, we can add multiples of 2pi to the principal value to get additional solutions.
The values of x in the interval (6pi, 8pi) for cos(x) = √(1/4) are:
x = Pi/3 + 2pi, Pi/3 + 4pi, Pi/3 + 6pi, ...
Case 2: cos(x) = -√(1/4)
Using the same process as in Case 1, we find the principal value x = (2pi - Pi/3).
Adding multiples of 2pi, we get additional solutions in the interval (6pi, 8pi):
x = (2pi - Pi/3) + 2pi, (2pi - Pi/3) + 4pi, (2pi - Pi/3) + 6pi, ...
To summarize, the solutions of the equation 4(cos(x)^2) - 1 = 0 in the interval (6pi, 8pi) are:
(Pi/3 + 2pi), (Pi/3 + 4pi), (Pi/3 + 6pi), ..., ((2pi - Pi/3) + 2pi), ((2pi - Pi/3) + 4pi), ((2pi - Pi/3) + 6pi), ...