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

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Answer 1 · 2018-05-07T02:37:03

#color(purple)(=> pi#

#sin^2x = 2 cos x + 2#

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

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

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

#(cos x = 1)^2 = 0#

#cos x + 1 = 0#

#cos x = -1 #

#color(purple)(x = cos ^-1 -1 = cos^-1 cos (pi) = pi#