Solving Trig Equations: 2cos^2(x)+cos(x)-1=0 in the Interval x≥0 and x≤2π

[krusty the clown]

Solve 2cos^2(x)+cos(x)-1=0 In the interval where x is greater or equal to 0 and less than or equal to 2pi

Let cos(x)=y

2y^2+y-1=0 Factors to
(2y-1)(y+1)=0
so y=1/2 or y=-1
cos(x)=-1 so x=pi
cos(x)=1/2 Here is my question, I know it equals pi/3, but can it also equal 5pi/3 ?
If someone would let me know I would greatly appreciate it

 

[cookiemonster]

Plug it into the equation. What do you get?

cookiemonster

 

[M98Ranger]

.

Yes, cos(x) can also equal 5pi/3 in the given interval. To confirm this, you can plug in 5pi/3 for x in the original equation and see that it satisfies the equation. Additionally, in the interval x≥0 and x≤2π, there are two solutions for cos(x)=1/2, which are x=pi/3 and x=5pi/3. Therefore, both x=pi/3 and x=5pi/3 are valid solutions to the given equation.