# Thread: [SOLVED] solving a trig. equation.

1. ## [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.

2. 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 $x = \frac{3\pi}{2}$

3. 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.
$1 + \cos{x} = 1 - \cos{x}$

$2\cos{x} = 0$

$\cos{x} = 0$

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

$x = \frac{3\pi}{2}$

4. thanks to both of you!