How do you solve #2sin^2 x-11cosx - 11 = 0# from 0 to 2pi?

1 Answer
Alan P.
Aug 5, 2015

#x=pi#

Explanation:

Since #sin^2(x) = 1-cos^2(x)#

#2sin^2(x) - 11 cos(x) - 11 = 0#
#color(white)("XXXX")#is equivalent to
#2-2cos^2(x) -11cos(x) - 11 =0#

#color(white)("XXXX")#simplifying
#2cos^2(x)+11cos(x)+9 = 0#

#color(white)("XXXX")#factoring
#(2cos(x)+9)(cos(x)+1) = 0#

#2cos(x)+9 = 0# or #cos(x)+1 =0#

#2cos(x)+9=0# is extraneous since it implies #cos(x)# is outside the range #[-1,+1]#

#cos(x)+1 = 0#
#color(white)("XXXX")##rarr cos(x) = -1#

within the range #x epsilon [0, 2pi]#
#color(white)("XXXX")##rarr x = pi#

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