How do you find all solutions of the equation #sin^2x+sinx=0#?

1 Answer
Jun 9, 2015

Answer:

#S={x|x=kpivvx=-3/2pi+2kpi ^^kinZZ}#

Explanation:

#sin^2x+sinx=0#
can be rewritten:
#sinx(sinx+1)=0#
So we can state that #sinx=0# or #sinx=-1# because multiplying two numbers can result zero if and only if one of the two is zero.
So let's solve the two parts of the equation:
1. #sin(x)=0#
It is basically #x=kpi# where #kinZZ#
2. #sin(x)=-1#
It is basically #x=-3/2pi+2kpi# where #kinZZ#

So the solutions are:
#S={x|x=kpivvx=-3/2pi+2kpi ^^kinZZ}#