# Trigonometric Equation 2

1. Jan 8, 2008

### TbbZz

1. The problem statement, all variables and given/known data
tan^2x = 2tanx*sinx

2. Relevant equations
N/A

3. The attempt at a solution

tan^2x = 2tanx*sinx

(tan^2x)/tanx = (2tanx*sinx)/tanx

tanx = 2sinx

sinx/cosx = 2sinx

cosx = 1/2

x = $$pi$$/3

$$\alpha$$ = $$pi$$/3 and 5$$pi$$/3

Not sure what I'm doing wrong.

The answer is $$pi$$/3 and 5$$pi$$/3 and 0, and $$pi$$

Thanks for the assistance, I greatly appreciate it.

2. Jan 8, 2008

### rock.freak667

Do not simply just divide by tanx
Bring 2tanx*sinx onto the other side and then factor out the tanx

Dividing by tanx implies that tanx is never equal to zero. Which it can be in this case.

3. Jan 8, 2008

### HallsofIvy

Staff Emeritus
Either tan2(x)/tan x= 2 tan(x) sin(x)/tan(x) or tan(x)= 0.

4. Jan 8, 2008

### TbbZz

Thanks for the help, rock.freak667 and HallsofIvy.

I got it now.