Question

in the interval (0,2pi) solve for x: cos^2x=cosx I NEED TO KNOW HOW TO DO THIS!

Answer 1

\[z^2=z\\
z^2-z=0\\
z(x-1)=0\\
z=0,z=1\]

Answer 2

cos(x) = 1,
When?

Answer 3

It is the x value on the unit circle,

Answer 4

\[z=\cos(x)\] here and then
\[\cos(x)=0\]\[\cos(x)=1\]

Answer 5

Then take the inverse to find x (technically the angle) \[x=\cos^{-1}(0)\]\[x=\cos^{-1}(1)\]

Answer 6

when is the x value either 0, or 1?

Answer 7

theta =