How do you solve 2/3 sin^2x-cosx=1 on the interval 90 to 180 degrees?

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Answer 1 · 2018-04-12T02:57:57

#x=120,180#

#2/3sin^2x-cosx=1#

#2/3(1-cos^2x)-cosx=1# Pythagorean identity #sin^2x+cos^2x=1#

#2/3-(2cos^2x)/3-cosx=1# expand brackets

#2-2cos^2x-3cosx=3# multiply both sides by #3#

#-1-2cos^2x-3cosx=0# subtract 3 from both sides

#2cos^2x+3cosx+1=0# multiply both sides by #-1#

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

#cosx=-1/2,-1#

#x=120, 180# for #[90,180]#