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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