How do you solve #sin2xcosx+sinx=0# in the interval [0,360]?

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Answer 1 · 2016-10-07T20:45:10

#:. S={0,pi,2pi}#

#sin2xcosx+sinx=0#

#2 sinxcosxcosx+sinx=0#

#sinx(2cos^2x+1)=0#

#sinx =0 or 2cos^2x+1=0#

#x=sin^-1 0# or #2cos^2x=-1->#Not possible since we cannot take a square root of a negative number.

#x=sin^-1 0 = 0,pi,2pi#

#:. S={0,pi,2pi}#