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

1 Answer

Aug 4, 2016

#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]#

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