can this be simplified further? 2cos²Θ + 2cosΘ - 3?

Question

the Θ symbol meaning theta

Answer 1

let x = cos A . . . . . . . if this is equated to zero
2 x^2 + 2 X - 3 = 0 . . . there is no exact factor . . root is not an integer
x = - 1.82288 . . . . and x = 0.82288
use x = 0.82288
cos A = 0.82288
A = 34 deg 37' 3"

Answer 2

Of course it can....

2cos²Θ + 2cosΘ - 3

2( cos²Θ + cosΘ - 3/2)

Now, let's factor cos²Θ + cosΘ - 3/2

Pretend that cosΘ = x
so we have
2( x² + x - 3/2)

We can now "factor" that polynomial. Easiest way of doing is that is as an earlier poster stated, use the quadratic forumula.

x² + x - 3/2 = 0

By use of the quadratic formula

x = (-1 +/- sqrt(1 + 6)) / 2
x = (-1 +/- sqrt(7)) / 2
so the roots are
x = (-1 + sqrt(7)) / 2
and
x = (-1 - sqrt(7)) / 2

This means that we can write
x² + x - 3/2

as
(x - (-1 + sqrt(7)) / 2)(x - (-1 - sqrt(7)) / 2)

or
(x + 1/2 - sqrt(7)/2)(x + 1/2 + sqrt(7) / 2)

Remember though that x was actually cosΘ

so
2cos²Θ + 2cosΘ - 3 ==
2 (cosΘ + 1/2 - sqrt(7)/2) (cosΘ+ 1/2 + sqrt(7) / 2)

It's ugly but simplified :)

Answer 3

Yes it can, but its awfully messy. Would be a lot less messy and still test your abilities if the question was 2cos²Θ + cosΘ - 3. You could then factor it as you would 2X² + X - 3.

Answer 4

No.