How do you solve #2sin^2x=cosx+1# in the interval #0<=x<=2pi#?

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Answer 1 · 2016-07-31T17:43:18

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

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

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

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

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

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

#cosx = 1/2 and cosx = -1#

#x = pi/3, (5pi)/3 and pi#

Hopefully this helps!