How do you solve #2cos^2x + cosx - 1= 0# and find the general solution?

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Answer 1 · 2016-04-23T09:04:42

#x = 2kpi+-pi/3, k+0, +-1, +-2, ...#and
x=an odd multiple of #pi=(2k+1)pi, k=0. +-1, +-2...#

This is a quadratic in cos x.

The roots are
#cos x =1/2 and -1#

The corresponding principal values of
arc cos (1/2) and arc cos(-1) are
#pi/3 and pi#.

The general values are

#x = 2kpi+-pi/3, k+0, +-1, +-2, ...#and

x=an odd multiple of #pi=(2k+1)pi, k=0. +-1, +-2...#

For the first root, there is a choice from #+-pi/3 or pi/3 and (5pi)/3#.

For the principal value. For the second, there is no choice.,