How do you solve #2cos2x + 2cosx-1=0#?

1 Answer
Alan P.
Feb 13, 2015

#2cos2x+2cosx−1=0# implies #x = arccos( (sqrt(13) - 1)/4)#
(Yes; it is ugly... but the point is how to get there...)

Remember the double angle formula
#cos(2x) = 2 cos^2(x) - 1#

So
#2cos^2x+2cosx−1=0#
becomes
# 4 cos^2 -2 +2 cos x - 1 = 0#
# 4 cos^2 x + 2 cos x - 3 = 0#

Using the standard formula for roots of a quadratic:
# (-b +- sqrt( b^2 - 4ac))/2a#

We get (with minor handling)
#cos(x) = ( (-1) +- sqrt(13))/4#

#x = arccos ( cos(x) )# by definition
noting, however, that cos(x) must fall in the range [-1, +1]
so the root #( (-1) - sqrt(13))/4# can be eliminated

Therefore
#x = arccos(( (-1) + sqrt(13))/4)#

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