Solve the following equations for all solutions of x

2cos(3x-1)=0

So do I just want to start out multiplying and getting 6cosx-2cos=0?

Question

Solve the following equations for all solutions of x

2cos(3x-1)=0

So do I just want to start out multiplying and getting 6cosx-2cos=0?

anonymous · Answer

not if it really is 
$$2\cos(3x-1)=0$$[

anonymous · Answer

cosine is a function. think if it as $f(x)=\cos(x)$ 
so you have $f(3x-1)=\cos(3x-1)$

anonymous · Answer

So you subtracted one of the cos(3x-1) from the other?

lgbasallote · Answer

you actually divide both sides by 2

lgbasallote · Answer

think of cos (3x - 1) as a variable y

so you have 2y = 0

to get y alone, you divide both sides by 2

make sense?

anonymous · Answer

Yeah...so, do I end up with cos(3x-1)=0

anonymous · Answer

Think of this as cosA=0 and solve for A

anonymous · Answer

Ugg...this class is terrible!

anonymous · Answer

If cos A = 0 then A is 90 degees, 270 degrees and so on or pi/2 or 3pi/2 etc

anonymous · Answer

I really appreciate the help.  Thanks everyone!

anonymous · Answer

Hope you do okay on this topic. Cheers for the medal

anonymous · Answer

Yeah...I've just been working n these all day and my mind is mush.  Thanks again.