How do you solve #cos 2x + 3 sinx - 2= 0#?

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Answer 1 · 2016-04-18T16:09:35

#S={pi/6 +2pin, (5pi)/6 +2pin, x=pi/2+2pin}#

Use Double Argument Property:
#cos2A=1-2sin^2A#

#1-2sin^2x+3sinx-2=0#

#2sin^2x-3sinx+1=0#

#(2sinx-1)(sinx-1)=0#

#2sinx-1=0 or sinx-1=0#

#sinx=1/2 or sinx =1#

#x=sin^-1(1/2) or x=sin^-1 1#

#x=pi/6 +2pin, (5pi)/6 +2pin or x=pi/2+2pin#

#S={pi/6 +2pin, (5pi)/6 +2pin, x=pi/2+2pin}#