How do you solve #2cos(x)sin(x)+sin(x)=0#?

1 Answer
Alan P.
Jul 14, 2015

#x= 0^o or 120^o or 180^o or 240^o#
or
#x= 0 or (2pi/3) or pi or (4pi)/3# radians
#color(white)("XXXX")#assuming we are restricted to #[0,360^o)#

Explanation:

If #2cos(x)sin(x)+ sin(x) = 0#
then
#color(white)("XXXX")##(sin(x))*(2cos(x)+1) = 0#

#rArr sin(x) = 0#
#color(white)("XXXX")##rarr x = 0^o or 180^o#
or
#rArr 2cos(x)+1 = 0#
#color(white)("XXXX")##rarr cos(x) = -1/2#

#color(white)("XXXX")##rarr x= 180^o +- 60^o#
#color(white)("XXXX")##color(white)("XXXX")#(standard trigonometric triangle)
#color(white)("XXXX")##rarr x = 120^o or 240^o#

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