How do you solve #2cos^2(3x)+5cos(3x)-3=0#?

1 Answer
Alan P.
Aug 1, 2016

For #x in [0,pi]#
#color(white)("XXX")color(green)(x=pi/9)#

Explanation:

Let #theta=3x#
and #k=cos(theta)#

The given equation becomes
#color(white)("XXX")2k^2+5k-3=0#

which can be factored as
#color(white)("XXX")(2k-1)(k+3)=0#

with possible solutions
#color(white)("XXX")k=1/2color(white)("XX")"or"color(white)("XX")k=-3#

but since #k=cos(theta)#
#color(white)("XXX")k in [-1,+1]#
and #k=-3# is an extraneous solution.

#k=cos(theta)=1/2# (for #theta in [0,pi]#)
#rarr theta =pi/3#

and since #3x=theta#
#rarr x=pi/9#

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