Given that http://mapleta7.eps.surrey.ac.uk:808...dllcdoeonn.gif is a root of the equation

http://mapleta7.eps.surrey.ac.uk:808...pngbjehook.gif,

find the other two roots of http://mapleta7.eps.surrey.ac.uk:808...ejjcekgdon.gif.

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- October 7th 2012, 09:06 AMTweetytrig equation,
Given that http://mapleta7.eps.surrey.ac.uk:808...dllcdoeonn.gif is a root of the equation

http://mapleta7.eps.surrey.ac.uk:808...pngbjehook.gif,

find the other two roots of http://mapleta7.eps.surrey.ac.uk:808...ejjcekgdon.gif.

- October 7th 2012, 09:40 AMskeeterRe: trig equation,

you know is a zero ... I advise using synthetic division to find the other two possible roots

btw ... check my algebra for any egregious sign/addition errors - October 8th 2012, 11:28 AMStefanTMRe: trig equation,
Hi Tweety,

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

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45 24 3 0 |1/3 - October 8th 2012, 12:23 PMSorobanRe: trig equation,
Hello, Tweety1

Quote:

We have: .

. . . . . . . .

. . . . . . . .

- October 8th 2012, 12:41 PMHallsofIvyRe: trig equation,
If sin x= 1/3, then so that we have

. 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. - October 8th 2012, 01:06 PMjohnsomeoneRe: trig equation,
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 .

Thus find the solutions to: .

(So , so or )