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

1 Answer
Alan P.
Jan 17, 2016

If #x in [-pi/2,+pi/2]# (the typical restriction for #"arcsin"#)
then #x=-pi/6# or #x=-pi/2#

Explanation:

#2cos^2x-3sinx=3#

#rarr 2(1-sin^2x)-3sinx-3=0#

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

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

#rarr (2sinx+1)(sinx+1)=0#

#rarr sinx=-1/2# or #sin x=-1#

#color(white)("XXXXXXXXXX")#assuming #x in [-pi/2,+pi/2]#
#rarr x=-pi/6# or #x=-pi/2#

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