2cos*2x-cosx=1 find all solutions to the equation on the domain [0,2ð]
i got ð/4, 7ð/4, 3ð/4, and 5ð/4 but only got two out of four points... what did i do wrong?
Is cos*2x supposed to mean cos^2 x?
Is 2ð supposed to mean 2 pi?
If so,
2 cos^2x - cos x -1 = 0
which factors into
(2 cos x +1)(cos x -1) = 0
The solutions occur where cos x = -1/2 or +1
If cos*2x means cos(2x), then
2[2 cos^2 x -1) - cos x -1 = 0
4 cos^2 x -cos x -3 = 0
(4 cosx +3)(cosx - 1) = 0
x is where cos x = -3/4 or +1
To solve the equation 2cos^2(x) - cos(x) = 1 on the domain [0, 2π], let's break it down step-by-step:
Step 1: Rewrite the equation
2cos^2(x) - cos(x) - 1 = 0
Step 2: Factor the equation
(2cos(x) + 1)(cos(x) - 1) = 0
Step 3: Set each factor equal to zero and solve for x
2cos(x) + 1 = 0 or cos(x) - 1 = 0
For the first factor:
2cos(x) = -1
cos(x) = -1/2
Using the unit circle, we know that cos(x) = -1/2 at π/3 and 5π/3 radians on the interval [0, 2π].
For the second factor:
cos(x) = 1
Using the unit circle, we know that cos(x) = 1 at 0 and 2π radians.
Therefore, the solutions on the interval [0, 2π] are x = 0, π/3, 5π/3, and 2π.
It seems like you made a mistake with one of the solutions. The correct solutions are x = 0, π/3, 5π/3, and 2π.
To find all the solutions to the equation 2cos^2(x) - cos(x) = 1 on the domain [0, 2π], we can rewrite the equation in terms of the cosine function:
2cos^2(x) - cos(x) - 1 = 0
Now, let's solve this quadratic equation for cos(x):
Multiply the equation by 2 to simplify it:
4cos^2(x) - 2cos(x) - 2 = 0
Rewrite the equation using the quadratic formula:
cos(x) = [-(-2) ± √((-2)^2 - 4(4)(-2))] / (2(4))
cos(x) = [2 ± √(4 + 32)] / 8
cos(x) = [2 ± √36] / 8
cos(x) = (2 ± 6) / 8
cos(x) = 8/8 or -4/8
cos(x) = 1 or -1/2
To find the solutions on the given domain [0, 2π], we need to determine the corresponding angles whose cosine values are 1 or -1/2.
For cos(x) = 1, the solutions in the given domain are x = 0 and x = 2π.
For cos(x) = -1/2, we need to find the angles whose cosine is -1/2. These angles occur in the second and third quadrants. Since the cosine function is positive in the fourth quadrant, we do not have any solutions in the fourth quadrant.
In the second quadrant, the angle that has a cosine of -1/2 is π + arccos(-1/2).
In the third quadrant, the angle that has a cosine of -1/2 is 2π - arccos(-1/2).
Therefore, the solutions in the given domain for cos(x) = -1/2 are x = π + arccos(-1/2) and x = 2π - arccos(-1/2).
Hence, the correct solutions on the domain [0, 2π] are:
x = 0, x = 2π, x = π + arccos(-1/2), and x = 2π - arccos(-1/2).
To calculate the values for π + arccos(-1/2) and 2π - arccos(-1/2), you can substitute the value of arccos(-1/2) into a calculator to find the precise values.