Solving 2 + cos2x = 3cosx: 0<x<2pi

[Karma]

The question is solve $$2 + cos2x=3cosx$$ for 0<- X <-2pi greater/equal to zero and less than or equal to 2pi.

My solution
$$2+cos2x=3cosx$$
$$2+cosx^2-sin^2=3cosx$$
$$2+cosx^2-(1-cosx^2)=3cosx$$
$$2+2cosx^2-1-3cosx=0$$
$$2cos^2-3cosx+1=0$$

Factoring
$$2x^2-3x+1=0$$
$$(2x-1)(x-1)=0$$
$$x=1 or x=1/2$$

solving for cos i get
$$cos(x)=1 --> x=0$$
$$cos(x)=1/2 -->x= pi/3 or 5pi/3$$

Did i do this Right?

 

[Hurkyl]

What about 2pi?


You have the right idea; I will complain that you used the same letter x for multiple things, though. The x in

2x² - 3 x + 1 = 0

is not the same as the x in

2 + cos 2x = 3 cos x.

You really should introduce a new variable when you do something like that, and you ought to indicate how it relates to the old variables. e.g.

Let y = cos x. Then, 2y² - 3 y + 1 = 0, ...



And you are missing one step (which is unfortunately missed a lot) -- now that you have some candidate solutions, you need to either plug them back into the original equation to see if they really are solutions, or you need to argue that all of your work is reversible.

If it's patently obvious you've used only reversible steps, then it's okay to skip this last one -- but otherwise you should do it.