How do you find all solutions of the equation in the interval (0, 2pi) #-cosx + 1 = 2sin^2x#?

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Answer 1 · 2016-04-19T17:50:09

#S={(2pi)/3,(4pi)/3}#

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

#2(1-cos^2x)+cosx-1=0#

#2-2cos^2x+cosx-1=0#

#2cos^2x-cosx-1=0#

#(2cosx+1)(cosx-1)=0#

#2cosx+1=0 or cosx-1=0#

#cosx=-1/2 or cosx=1#

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

#x=+- (2pi)/3 +2pin or x=0+2pin#

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

#n=1, x=(8pi)/3, (4pi)/3,2pi#

#S={(2pi)/3,(4pi)/3}#