# [SOLVED] solving a trig. equation.

• May 2nd 2009, 10:09 AM
brentwoodbc
[SOLVED] solving a trig. equation.
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.
• May 2nd 2009, 10:17 AM
running-gag
Quote:

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 $x = \frac{3\pi}{2}$
• May 2nd 2009, 10:18 AM
skeeter
Quote:

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.

$1 + \cos{x} = 1 - \cos{x}$

$2\cos{x} = 0$

$\cos{x} = 0$

$x = \frac{\pi}{2}$

$x = \frac{3\pi}{2}$
• May 2nd 2009, 10:19 AM
brentwoodbc
thanks to both of you!(Clapping)(Clapping)(Clapping)(Clapping)