Question

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

Answer 1

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`