How do I solve for 0° ≤ x ≤ 360° given sin 2x = - sin x?

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Answer 1 · 2018-02-21T19:35:07

#0^@ , 120^@ ,180^@ , 240^@ , 360^@#

Identity.

#color(red)bb(sin2x=2sinxcosx)#

#sin(2x)=-sinx#

#sin(2x)+sinx=0#

Using identity:

#2sinxcosx+sinx=0#

Factor:

#sinx(2cosx+1)=0#

#sinx=0#

#x=arcsin(sinx)=arcsin(0)=>x= 0^@, 180^@,360^@#

#2cosx+1=0#

#cosx=-1/2#

#x=arccos(cosx)=arccos(-1/2)=>x=120^@ , 240^@#