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

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Answer 1 · 2015-02-03T15:25:08

The answers are: #x=kpi# and #x=+-2/3pi+2kpi#.

Because, for the double-angle formula of sinus, that says:

#sin2alpha=2sinalphacosalpha#, the equation becomes:

#2sinxcosx+sinx=0rArrsinx(2cosx+1)=0rArr#

#sinx=0rArrx=kpi#;

and

#2cosx+1=0rArrcosx=-1/2rArrx=+-2/3pi+2kpi#.