# Trig identies

1. Jul 20, 2009

### kavipach

1. The problem statement, all variables and given/known data
2cos2x-2cosx...how do you simplify this further?

2. Relevant equations

3. The attempt at a solution
2(cos2x-cosx)..but i have to find 0=2cos2x-2cosx so this doesnt really help me.

2. Jul 21, 2009

### Office_Shredder

Staff Emeritus
You're trying to find when 2cos(2x)-2cos(x)=0? Start by what you did... you want to find x such that cos(2x)=cos(x). Hint: cosine is periodic

3. Jul 21, 2009

### kavipach

would it be pi/2 and 3pi/2?

4. Jul 21, 2009

### queenofbabes

nope....maybe try drawing cos(2x) and cos(x)?

5. Jul 21, 2009

### zcd

$$\cos(2x)=2\cos^{2}(x)-1$$, so $$2\cos(2x)-2\cos(x)=2\cos^{2}(x)-1\cos(x)-1=0$$. Factoring yields $$(2\cos(x)+1)(\cos(x)-1)=0$$