[Note: I've tried this problem 4 times already and still have it wrong. I know the steps to doing it but for some reason its wrong. an someone please help me get these answers for (b).]
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 equation.
My Answer--> θ= (π+6nπ)/6 , (5π+6nπ)/6
(Correct)
(b) Find the solutions in the interval [0, 2π).
θ=____________________________
[Note: my answer was wrong.
π/2,π/6,3π/2,7π/6]
2 cos 2θ − 1 = 0
cos 2Ø = 1/2
I know cos π/3 = 1/2 , and the cosine is positive in I and IV
so 2Ø = π/3 , in I
OR
2Ø = 2π - π/3 = 5π/3 , in IV
Ø = π/6 or Ø = 5π/6
(Ø = 30° or Ø = 150°)
but the period of cos 2Ø is 2π/2 or π
so adding/subtracting multiples of π to any answer will yield a new answer
general solution:
π/6 + kπ or 5π/6 + kπ , where k is an integer
solutions in [0,2π] :
π/6, 7π/6, 5π/6, 11π/6
or
in degrees: 30°, 210°, 150°, 330°
verify my answers by using a calculator, they work.
Thank you so much for your help!!!
You're so awesome!!! :D
To solve the equation 2 cos 2θ − 1 = 0 in the interval [0, 2π), you need to follow these steps:
1. Start by isolating the cosine term by adding 1 to both sides of the equation:
2 cos 2θ = 1
2. Divide both sides by 2 to get rid of the coefficient of the cosine term:
cos 2θ = 1/2
3. Now, find the angles whose cosine value is equal to 1/2. This can be done using the inverse cosine function (also known as arccosine or cos^(-1)). The inverse cosine of 1/2 is π/3 in radians. However, note that the cosine function is positive in the first and fourth quadrants, so you also need to consider the reference angle in the fourth quadrant, which is 2π/3.
4. To account for the multiple possible solutions, we use the periodicity of the cosine function. Since the period of cosine is 2π, we can add integer multiples of 2π to the solution angles.
Putting it all together, the solutions in the interval [0, 2π) are:
θ = π/3 + 2nπ, where n is an integer, and
θ = 2π/3 + 2nπ, where n is an integer.
Therefore, the correct answer to part (b) is:
θ = π/3, 2π/3.
Your answers of π/2, π/6, 3π/2, and 7π/6 were incorrect because they do not satisfy the equation 2 cos 2θ − 1 = 0.