Given that is a root of the equation
find the other two roots of .
you can substitue sin(x) =t
having P(t) = 45t^3 + 9t^2 - 5t -1 = 0
You can divide with t-1/3 (polynomial division) receiving as quotient (without rest) a polynom of second grad, resolvable with pq- or abc-Formel
Alternatively you can use the Horner schema to get the quotient:
45 9 -5 -1
45 24 3 0 |1/3
. Multiplying both sides by 3, 40+ 24=64= 8(3+ 5),
(I point this out because the first time I tried this I made a silly error and thought it wasn't true!)
Lettying y= sin(x), this equation becomes so that . Yes, y= 1/3 satisfies that so that y- 1/3 is a factor. Using either polynomial division or "synthetic division" shows that the other factor is so we can solve the quadratic equation by factoring.
Once you know one root of a cubic, there's a way to find the other two roots w/o polynomial long division. However, doing polynomial division isn't that difficult.
After division, have lead coef 1: Let have a root . Let the other roots be .
Since the product of all 3 roots is , and the sum of three roots is , have:
and (obviously provided ).
Thus are the solutions to: , so
are the solutions to: , so
are the solutions to: .
Note that polynomial long division will give you the same result.
In this case, using the others have computed here, you'd get
, so .
The above becomes: .
Thus find the solutions to: .
(So , so or )