Recall:
[tex] { \sin(x) }^{2} + { \cos(x) }^{2} = 1[/tex]
[tex] \sin(x) \cos(x) = 0 \\ \sin(x) = 0 \: \: or \: \cos(x) = 0[/tex]
If sin(x) = 0,
[tex] { \sin(x) }^{2} + { \cos(x) }^{2} = 1 \\ {0}^{2} + { \cos(x) }^{2} = 1 \\ { \cos(x) }^{2} = 1 \\ \cos(x) = 1 \\ x = 0[/tex]
If cos(x) = 0,
[tex] { \sin(x) }^{2} + { \cos(x) }^{2} = 1 \\ { \sin(x) }^{2} + { 0 }^{2} = 1 \\ { \sin(x) }^{2} = 1 \\ \sin(x) = 1 \\ x = 90[/tex]
Therefore, x is either 90° or 0°.
