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

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

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