Question

How do you solve `sinxcosx-2cosx=0`?

Answer 1

`x in {pi/2+kpi, k in ZZ}`

Explanation:

`sin(x) cos(x) - 2cos(x)=0`

`rarr color(red)(cos(x)) * color(blue)(""(sin(x)-2)) = 0`

`rarr color(red)(cos(x)=0) color(white)("XXX")"or"color(white)("XXX") color(blue)(sin(x)-2=0`

`rarr color(red)(x=pi/2+k * pi)" or "color(white)("XX")color(blue)(sin(x)=2`

`color(white)("XXXXXXXXXXXXXXXX")color(blue)("impossible since " sin in [-1,+1]`