[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

(b) Find the solutions in the interval [0, 2π).


[Note: my answer was wrong.

  1. 0
  2. 3
asked by Amber
  1. 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
    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


    in degrees: 30°, 210°, 150°, 330°

    verify my answers by using a calculator, they work.

    1. 0
    posted by Reiny
  2. Thank you so much for your help!!!
    You're so awesome!!! :D

    1. 0
    posted by Amber

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