Find all solutions to the given equation in the interval [0,2π). Give the exact solution, including "pi" for π. For any unused answer boxes, enter DNE in all capital letters.
(a) 2cos x=2
so cos x=1 =0..(now what?)
(b) 4cos x+2=0
cos x = -1/2 (120)
cos is negative in quads 2 and 3
not sure what Iḿ doing..thank you
help please...I donẗ get this..
2cosx = 2
cosx = 1
you should familiarize yourself with the basic sine and cosine curves,
looking at the cosine curve notice that when x = 0 , cos0 = 1
and when x = 2π, cos (2π) = 1
so your solution is x = 0, 2π for your required interval.
if all else fails, get your calculator, set it to Rad and enter
2ndF, cos, 1 , =
and you will get 0, so x = 0 is a solution
you know the period of the cosine curve is 2π, so 0+2π will be your
next solution.
(b) 4cos x+2=0
cos x = -1/2, you are right to say x must be in quads II and III
you should also know that cos 60° = +1/2 , but we have -1/2
so in quad II, your x must be 180-60 or 120°
in quad III, your x must be 180+60 or 240°
in radians that would be 2π/3 and 4π/3
so x = 2π/3, 4π/3
No problem! I'll be happy to help you step by step.
(a) For the equation 2cos(x) = 2, we want to find all the values of x in the interval [0, 2π) that satisfy this equation.
To solve for x, we need to isolate cos(x) by dividing both sides of the equation by 2:
cos(x) = 1/2.
The values of cos(x) that equal 1/2 are π/3 and 5π/3 in the interval [0, 2π).
To find these values, you can use the unit circle or the inverse cosine function (arccos):
x = π/3 and x = 5π/3.
So, the exact solutions to the equation 2cos(x) = 2 in the interval [0, 2π) are x = π/3 and x = 5π/3.
(b) For the equation 4cos(x) + 2 = 0, we want to find all the values of x in the interval [0, 2π) that satisfy this equation.
To solve for x, we need to isolate cos(x) by subtracting 2 from both sides of the equation:
4cos(x) = -2.
Now, divide both sides of the equation by 4 to get cos(x) = -1/2.
The values of cos(x) that equal -1/2 are 2π/3 and 4π/3 in the interval [0, 2π).
Again, you can use the unit circle or the inverse cosine function (arccos) to find these values.
So, the exact solutions to the equation 4cos(x) + 2 = 0 in the interval [0, 2π) are x = 2π/3 and x = 4π/3.
I hope this clarifies the process for finding the solutions in these equations! Let me know if there is anything else I can help you with.
No problem, I'll be glad to help you with these equations.
(a) To solve the equation 2cos x = 2, we need to isolate the cosine term by dividing both sides of the equation by 2. This gives us cos x = 1.
Since cosine(x) = 1 when x = 0, we know that 0 is a solution in the interval [0, 2π).
In the unit circle, cosine value of 1 occurs at two places: at 0 radians (0°) and at 2π radians (360°), which are both in the given interval. So, we have x = 0 and x = 2π as solutions.
Therefore, the solutions to the equation in the interval [0, 2π) are x = 0 and x = 2π.
(b) Similarly, to solve the equation 4cos x + 2 = 0, we first subtract 2 from both sides to isolate the cosine term. This gives us 4cos x = -2.
Next, we divide both sides by 4 to obtain cos x = -1/2.
Since cosine is negative at 120° and 240°, which are in quadrants 2 and 3, respectively, we can write x = π/3 and x = 5π/3 as solutions.
Therefore, the solutions to the equation in the interval [0, 2π) are x = π/3 and x = 5π/3.
I hope this helps! Let me know if you have any further questions or if there's anything more I can assist you with.