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

1 Answer
Alan P.
Oct 21, 2015

#x in {color(red)(npi), color(blue)((2pi)/3+n(2pi)), color(green)((4pi)/3+n(2pi))} AA n in ZZ#

Explanation:

#sin(x)+2sin(x)cos(x) = 0#

Factoring the left side:
#(sin(x))*(1+2cos(x))=0#

Either #color(red)(sin(x) = 0)# or #color(orange)(1+2cos(x)=0)#

If #color(red)(sin(x)=0)#
#rarr x in {color(red)(npi)} AA n in ZZ#
(i.e. #x# is a multiple of #pi# radians)

If #color(orange)(1+2cos(x)=0)#
#rarr cos(x) = -1/2#
withing the interval #[0,2pi]#
#x= color(blue)((2pi)/3)# or #color(green)(x=(4pi)/3)#
or, more generally:
#x= color(blue)((2pi)/3+n*2pi)# or #color(green)(x=(4pi)/3+n*2pi) AAn in ZZ#

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