How do you find the general solutions for #cosx tanx - 2 cos^2x = -1#?

1 Answer
Alan P.
Aug 22, 2015

For #AAn in ZZ#
#color(white)("XXXX")x=pi/6+npi or (5pi)/6+npi or (3pi)/2+npi#

Explanation:

#cos(x)*tan(x) - 2cos^2(x) = -1#
#rarrcolor(white)("XXX")cancel(cos(x))*sin(x)/cancel(cos(x))-2 (1-sin^2(x)) = -1#

#rarrcolor(white)("XXX")2sin^2(x)+sin(x)-1 = 0#

#rarrcolor(white)("XXX")#(2sin(x)-1)*(sin(x)+1) = 0#

#rarrcolor(white)("XXX")sin(x) = 1/2 or sin(x)=-1#

Within #[0,2pi]#
#color(white)("XXX")sin(x)=1/2color(white)("XXX")rarrcolor(white)("XXX")x=pi/6 or x=(5pi)/6#

#color(white)("XXX")sin(x) = -1color(white)("XXX")rarrcolor(white)("XXX")x=(3pi)/2#

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