facto2 said:
My instructor said that my answer for cosx=1 is not completly correct. I cannot figure out why. Any help with this would be much appreciated. Thank you.
I assume your problem was:
Solve for 'x' when
\(\displaystyle \cos^2(x) \,- 1 = \, 0\)
then, the solution is:
\(\displaystyle \cos^2(x) \, - \, 1 \, = \, 0\)
\(\displaystyle [\cos(x) \, - \, 1] \cdot [\cos(x) \, + \, 1]\, = \, 0\)
\(\displaystyle \cos(x) \, + \, 1 \, = 0\)
..............or............... \(\displaystyle \cos(x) \, - \, 1 \, = \, 0\)
then
\(\displaystyle \cos(x) \, = \, \pm 1\)
Thus your answer was
partly correct.
