How do you solve #2sin^2x-cosx=1# on the interval [0,2pi]?

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Answer 1 · 2016-02-19T13:35:21

#{pi/3, pi, {5pi}/3}#

Use the identity

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

to make the equation into a quadratic equation w.r.t. #cosx#. Then proceed to solve the quadratic by factorization/completing the square.

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

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

#= 1#

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

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

This means that

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

#cosx# is positive for #0 < x < pi/2# #and# #{3pi}/2 < x < 2pi#.

#x = pi/3 or pi or {5pi}/3#