Solving A Trignometric Equation

[TrueStar]

Homework Statement

Find all solutions to the equation in the interval [0,2$$\pi$$] algebraically.

cos2x(2cos+1)=0

Homework Equations

NA

The Attempt at a Solution

This is what I've done, but I don't think it's right. I set cos2x=0 and 2cos+1=0. Since cos2x is a multi-angle, I let t=2x and rewrote it as cost=0 That means t equals $$\pi$$\2+n$$\pi$$.

That also means 2x=$$\pi$$\2+n$$\pi$$. After dividing out the 2, I get x=$$\pi$$\4+n$$\pi$$\2.

I get the answers $$\pi$$\4, 3$$\pi$$\4, 5$$\pi$$\4, and 7$$\pi$$\4. This doesn't seem right to me. Shouldn't I only get two answers.

The other one seems easier. I got cosx by itself and it is cosx=-1\2. That means x is equal to 2$$\pi$$\3 and 4$$\pi$$\3.

Am I doing this correctly?

Thanks!

 

[TrueStar]

The pi symbols aren't supposed to be raised up like that. It's just pi; sorry about that.

 

[Bohrok]

cosx=0 has two solutions, but you have four solutions to cos2x=0. That 2 multiplied with the x in cos2x makes its period only half as long as that of cosx, going from a period of 2$\pi$ to $\pi$
cosx=0 has just two solutions in [0, $\pi$], so cos2x=0 should have twice as many solutions, because it now goes through two periods in [0, $\pi$].

Hopefully this doesn't make it more complicated for you. You can also look at the graph of cos2x to see that there are four solutions in [0, $\pi$].


Instead of the [ tex ] tags, you can use [ itex ] where the LaTeX fits better within other text like you have. Also, your division lines should be going the other way. For example, $\pi$/4

 

[TrueStar]

Thank you for you reply! I plugged in the equations separately, and then as the original equation. It really helped visualize what was going on and I could double check using the trace function. I did not think to try this before. It all makes sense now!

I'll try to clean up my tags and symbols; thanks for the tips.

 

[HallsofIvy]

?? What does "2cos+1" mean? Is that supposed to be "2cos(x)+ 1"?
If the problem is to solve cos(2x)(2cos(x)+ 1)= 0 then, yes, either cos(2x)= 0 or 2cos(x)+ 1= 0. cosine is 0, in the interval $[0, 2\pi]$ only for 0 and $2\pi$ so you have 2x= 0 and $2x= 2\pi$ as roots.

2cos(x)+ 1= 0 gives cos(x)= -1/2. Think about dividing an equilateral triangle in half.

 

[TrueStar]

It's supposed to be 2cos(x)+1. I'm sorry, I mistyped it. I calculated the problem correctly and Bohrok helped me understand why x has twice as many solutions for cos(2x)=0.