[SOLVED] solving a trig. equation.

**brentwoodbc** · May 2nd 2009, 10:09 AM

for
0 less than or equal to x less than 2pi.

1+cosx=1-cosx?
I cant figure out how to solve this, you cant really factor it/ use the quadratic equation.
I assume it has something to do with the identity
cosx=cos(-x)

If anyone could steer me in the direction of solving this Id appreciate it, thank's.

**running-gag** · May 2nd 2009, 10:17 AM

Originally Posted by brentwoodbc

for
0 less than or equal to x less than 2pi.

1+cosx=1-cosx?
I cant figure out how to solve this, you cant really factor it/ use the quadratic equation.
I assume it has something to do with the identity
cosx=cos(-x)

If anyone could steer me in the direction of solving this Id appreciate it, thank's.

Hi

If I well understand your equation is $1+\cos x = 1-\cos x$ right ?

This is equivalent to $2\cos x = 0$ therefore $x = \frac{\pi}{2}$ or $x = \frac{3\pi}{2}$

**skeeter** · May 2nd 2009, 10:18 AM

Originally Posted by brentwoodbc

for
0 less than or equal to x less than 2pi.

1+cosx=1-cosx?
I cant figure out how to solve this, you cant really factor it/ use the quadratic equation.
I assume it has something to do with the identity
cosx=cos(-x)

If anyone could steer me in the direction of solving this Id appreciate it, thank's.