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.
[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 answers
Thank you so much for your help!!!
You're so awesome!!! :D
You're so awesome!!! :D
1 answer