How do you solve #cos^2 x-sin^2 x=sin x# over the interval #-pi<x<=pi#?

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Answer 1 · 2015-04-16T00:59:33

Write: #cos^2(x)=1-sin^2(x)# so that you get:

#1-sin^2(x)-sin^2(x)-sin(x)=0#
#-2sin^2(x)-sin(x)+1=0#

Set: #sin(x)=t# and get:

#-2t^2-t+1=0#
#t_1=-1#
#t_2=1/2#

but #sin(x)=t#

And in your interval:

#sin(x)=-1# and #x=-pi/2#
#sin(x)=1/2# and #x=pi/6 and x=5pi/6#

Hope it helps.