SOLUTION: Find the solutions of the equation that are in the interval [0, 2*Pi); separate your answers with commas. - 2cos 2(x)+ cos(x)+ 1 = 0; I already found 2Pi/3 and 4Pi/3 as answers

Algebra ->  Trigonometry-basics -> SOLUTION: Find the solutions of the equation that are in the interval [0, 2*Pi); separate your answers with commas. - 2cos 2(x)+ cos(x)+ 1 = 0; I already found 2Pi/3 and 4Pi/3 as answers      Log On


   

Question 183369This question is from textbook
: Find the solutions of the equation that are in the interval [0, 2*Pi); separate your answers with commas.
- 2cos 2(x)+ cos(x)+ 1 = 0;
I already found 2Pi/3 and 4Pi/3 as answers and they came up correct but the system says there are more. I am having trouble finding the others. Any help would be greatly appreciated. This question is from textbook

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
-2%28cos%28x%29%29%5E2%2Bcos%28x%29%2B1=0 Start with the given equation.


Notice how the term "cos(x)" is used most in this equation. So let z=cos%28x%29


-2z%5E2%2Bz%2B1=0 Plug in z=cos%28x%29


-%282z%2B1%29%28z-1%29=0 Factor the left side.


2z%2B1=0 or z-1=0 Set each factor equal to zero


Now let's solve each individual equation

------------------------------

2z%2B1=0 Start with the first equation


2cos%28x%29%2B1=0 Plug in z=cos%28x%29


cos%28x%29=-1%2F2 Isolate the term "cos(x)"


x=2pi%2F3 or x=4pi%2F3 Solve for "x". Since we're only concerned about the interval [), this means that there are only two solutions.

------------------------------


z-1=0 Move onto the next equation


cos%28x%29-1=0 Plug in z=cos%28x%29


cos%28x%29=1 Add 1 to both sides.


x=0 Solve for "x". Note: because we're limited to the interval [), there's only one solution




===============================

Answer:

So on the interval [), we have the three solutions: x=0, x=2pi%2F3 or x=4pi%2F3