SOLUTION: Find the solutions of the equation that are in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.) cos 2u &#8722

Algebra ->  Trigonometry-basics -> SOLUTION: Find the solutions of the equation that are in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.) cos 2u &#8722      Log On


   

Question 963379: Find the solutions of the equation that are in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
cos 2u − cos u = 0
Found 2 solutions by lwsshak3, ikleyn:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the solutions of the equation that are in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
cos 2u − cos u = 0
***
cos 2u − cos u = 0
cos^2(u)-sin^2(u)-cosu=0
cos^2(u)-1+cos^2(u)-cosu=0
2cos^2(u)-cosu-1=0
(2cosu+1)(cosu-1)=0
..
2cosu+1=0
cosu=-1/2
u=π/3, 5π/3
or
cosu-1=0
cosu=1
u=0

Answer by ikleyn(53875) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the solutions of the equation that are in the interval [0, 2π).
(Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
cos 2u − cos u = 0
~~~~~~~~~~~~~~~~~~~~~~~


        The solution and the answer in the post by @lwsshak3 both are incorrect.
        I came to bring a correct solution.


cos(2u) − cos u = 0
cos^2(u) - sin^2(u) - cosu = 0
cos^2(u) - 1 + cos^2(u) - cosu = 0
2cos^2(u) - cosu - 1 = 0
(2cosu+1)(cosu-1)=0
..
2cos(u)+1 = 0
cos(u) = -1/2
u = 2π/3, 4π/3
or
cos(u) - 1 = 0
cos(u) = 1
u = 0.

Solved correctly.