trig equation,

**Tweety** · October 7th 2012, 08:06 AM

Given that

is a root of the equation

,
find the other two roots of

.

**skeeter** · October 7th 2012, 08:40 AM

$40\sin{x} + 8 = 9\cos^2{x}(1 + 5\sin{x})$

$40\sin{x} + 8 = 9(1 - \sin^2{x})(1 + 5\sin{x})$

$40\sin{x} + 8 = 9(1 + 5\sin{x} - \sin^2{x} - 5\sin^3{x})$

$40\sin{x} + 8 = 9 + 45\sin{x} - 9\sin^2{x} - 45\sin^3{x}$

$45\sin^3{x} + 9\sin^2{x} - 5\sin{x} - 1 = 0$

you know $\sin{x} = \frac{1}{3}$ 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

**StefanTM** · October 8th 2012, 10:28 AM

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
_______________________

45 24 3 0 |1/3

**Soroban** · October 8th 2012, 11:23 AM

Hello, Tweety1

$\text{Given: }\:\sin x = \tfrac{1}{3}\,\text{ is a root of the equation: }\:40\sin x + 8 \;=\;9(1+5\sin x)\cos^2\!x$

$\text{Find the other two roots of }\sin x.$

We have: . $8(5\sin x + 1) \:=\:9(1+5\sin x)\cos^2\!x$

. . . . . . . . $8(5\sin x + 1) - 9(5\sin x + 1)\cos^2\!x \:=\:0$

. . . . . . . . $(5\sin x + 1)(8 - 9\cos^2\!x) \:=\:0$

$5\sin x + 1 \:=\;0 \quad\Rightarrow\quad \boxed{\sin x \:=\:\text{-}\tfrac{1}{5}}$

$8-9\cos^2\!x \:=\:0 \quad\Rightarrow\quad \cos^2\!x \:=\:\tfrac{8}{9}$
$\sin^2\!x \:=\:1-\cos^2\!x \quad\Rightarrow\quad \sin^2\!x \:=\:1 - \tfrac{8}{9} \:=\:\tfrac{1}{9} \quad\Rightarrow\quad \boxed{\sin x \:=\:\pm\tfrac{1}{3}}$

**HallsofIvy** · October 8th 2012, 11:41 AM

Originally Posted by Tweety

Given that

is a root of the equation

,
find the other two roots of

.

If sin x= 1/3, then $cos^2(x)= 1- sin^2(x)= 1- 1/9= 8/9$ so that we have
$40sin x+ 8= 9cos^2(x)(1+ 5sin(x))= \frac{40}{3}+ 8= 8(1+ 5/3)$ . 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 $40y+ 8= 9(1- y^2)(1+ 5y)= 9(1+ 5y- y^2- 5y^3)= 9+ 45y- 9y^2- 45y^3$ so that $45y^3+ 9y^2- 5y- 1= 0$ . 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 $45y^2+ 24y+ 3$ so we can solve the quadratic equation $45y^2+ 24y+ 3= 0$ by factoring.

**johnsomeone** · October 8th 2012, 12:06 PM

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 $p(x) = x^3 + a_2x^2 + a_1x + a_0$ have a root $r_0$ . Let the other roots be $r_1, r_2$ .

Since the product of all 3 roots is $-a_0$ , and the sum of three roots is $-a_2$ , have:

$r_1 + r_2 = -a_2-r_0$ and $r_1 r_2 = \frac{-a_0}{r_0}$ (obviously provided $r_0 \ne 0$ ).

Thus $r_1, r_2$ are the solutions to: $(y - r_1)(y-r_2) = y^2 - (r_1 + r_2)y + (r_1 r_2) = 0$ , so

$r_1, r_2$ are the solutions to: $y^2 - (-a_2-r_0)y + \left(\frac{-a_0}{r_0}\right) = 0$ , so

$r_1, r_2$ are the solutions to: $y^2 + (a_2 + r_0)y - \left(\frac{a_0}{r_0}\right) = 0$ .

Note that polynomial long division will give you the same result.

--------------------------

In this case, using the $45\sin(t)^3 + 9\sin(t)^2 - 5\sin(t) -1 = 0$ others have computed here, you'd get

$\sin(t)^3 + (9/45)\sin(t)^2 - (5/45)\sin(t) - (1/45) = 0$ , so $a_2 = 1/5, a_0 = - 1/45, r_0 = 1/3$ .

The above becomes: $\sin^2(t) + ((1/5) + (1/3)) \sin(t) - \left(\frac{(-1/45)}{(1/3)}\right) = 0$ .

Thus $\sin^2(t) + (8/15) \sin(t) + (1/15) = 0$ .

Thus find the solutions to: $15\sin^2(t) + 8 \sin(t) + 1 = 0$ .

(So $(5\sin(t) + 1)(3\sin(t) + 1) = 0$ , so $\sin(t) = -1/5$ or $\sin(t) = -1/3$ )

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