How do you solve #2cos^2x=3sinx#?

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Answer 1 · 2016-11-22T19:41:30

#x=pi/6+2pin#
#x=5pi/6+2pin#

#2cos^2x=3sinx#

Rewrite #cos^2x# as #1-sin^2x# because of the identity #sin^2x+cos^2x=1#:

#2(1-sin^2x)=3sinx#

Distribute the left side and let one side be zero:
#2-2sin^2x-3sinx=0#

#-2sin^2x-3sinx+2=0#

Factor:
#-2sin^2x-4sinx+sinx+2=0#
#-2sinx(sinx+2)+(sinx+2)=0#
#(-2sinx+1)(sinx+2)=0#

Use zero product rule:

#sinx+2cancel=0#
#sinxcancel=-2#

#-2sinx+1=0#
#-2sinx=-1#
#sinx=1/2#

#x=pi/6+2pin#
#x=5pi/6+2pin#