Find all solutions of the equation in the interval
, 02π
.
=2sin2x−2cosx
Write your answer in radians in terms of
π
.
If there is more than one solution, separate them with commas.
I assume you mean
2sin2x - 2cosx = 0
4sinx cosx - 2cosx = 0
2cosx(2sinx-1) = 0
so,
cosx = 0
or
sinx = 1/2
so, there will be four solutions in [0,2π)
To find the solutions of the equation 2sin^2x - 2cosx in the interval [0, 2π], we can start by factoring out a common factor of 2:
2(sin^2x - cosx) = 0
Next, we apply the identity sin^2x + cos^2x = 1:
2(1 - cosx - cosx) = 0
Simplifying further:
2(1 - 2cosx) = 0
Now, we solve for cosx:
1 - 2cosx = 0
2cosx = 1
cosx = 1/2
To find the solutions in the given interval [0, 2π], we can examine the unit circle for angles where cosx = 1/2. These angles occur at π/3 and 5π/3.
So, the solutions of the equation in the interval [0, 2π] are:
x = π/3, 5π/3
To find the solutions of the equation in the interval [0, 2π], we need to solve the equation and then express the solutions in radians in terms of π.
The given equation is:
2sin^2(x) - 2cos(x) = 0
To solve for x, we can use trigonometric identities to rewrite sin^2(x) in terms of cos(x):
sin^2(x) = (1 - cos^2(x))
Substituting this back into the equation:
2(1 - cos^2(x)) - 2cos(x) = 0
Expanding:
2 - 2cos^2(x) - 2cos(x) = 0
Simplifying:
2cos^2(x) + 2cos(x) - 2 = 0
Dividing through by 2:
cos^2(x) + cos(x) - 1 = 0
Now, we have a quadratic equation in terms of cos(x). We can solve this by factoring or using the quadratic formula.
Let's solve it using the quadratic formula:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 1, and c = -1. Plugging these values into the quadratic formula, we get:
cos(x) = (-1 ± √(1^2 - 4(1)(-1))) / (2(1))
cos(x) = (-1 ± √(1 + 4)) / 2
cos(x) = (-1 ± √5) / 2
Now, we have two possible values for cos(x):
1) cos(x) = (-1 + √5) / 2
2) cos(x) = (-1 - √5) / 2
To find the corresponding values of x, we need to take the inverse cosine (cos^-1) of these values:
1) x = cos^-1((-1 + √5) / 2)
2) x = cos^-1((-1 - √5) / 2)
However, we have to remember the interval [0, 2π]. So we need to find the values of x that lie within this interval.
Let's calculate the values of x using a calculator or software:
1) x ≈ 0.6223 radians
2) x ≈ 2.5180 radians
However, we need to express the solutions in terms of π. We can do this by dividing the values of x by π:
1) x/π ≈ 0.6223/π
2) x/π ≈ 2.5180/π
Thus, the solutions of the equation in the interval [0, 2π] are:
x ≈ (0.6223/π)π
x ≈ (2.5180/π)π
Simplifying:
x ≈ 0.6223π
x ≈ 2.5180π
Therefore, the solutions in radians in terms of π are approximately:
x ≈ 0.6223π, 2.5180π