Question

how do you solve cos^2 x−cosx=0?

Answer 1

\[\cos^2(x)-\cos(x)=0\]
\[\cos^2(x)=\cos(x)\]
*divide both sides by cos(x)
\[\cos(x)=1\]
turn 1 into cos with cos^-1 or arccos or shift cos, however you would like to call it.
\[\cos(x)=\cos(0)\]
You now can equate angles
\[x=0+360k\]
\[x=360k\]

Answer 2

*About the division by cos(x) above, when you divide by a variable you must state it equals to zero, there lies your second answer:
\[\cos(x)=0\]
Same deal, turn 0 into cos:
\[\cos(x)=\cos(90)\]
Equate angles:
\[x=90+180k\]

Answer 3

The answers:
\[x_1=360k\]
\[x_2=90+180k\]
The addition of 360k and 180k just takes into account that x has infinite solutions every 360 degrees