How do you solve #cos^2 x-sin^2 x=sin x#; for #-pi<x<=pi#?

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Answer 1 · 2016-04-19T18:17:21

#S={-pi/2, pi/6,(5pi)/6}#

Use Property: # cos^2x+sin^2x=1#

#1-sin^2x-sin^2x=sinx#

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

#2sin^2x+sinx-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 (3pi)/2 +2pin#

#n=-1, x=-(11pi)/6,-(7pi)/6,-pi/2#

#n=0, x=pi/6, (5pi)/6, (3pi)/2#

#S={-pi/2, pi/6,(5pi)/6}#