Given that is a root of the equation
,
find the other two roots of .
$\displaystyle 40\sin{x} + 8 = 9\cos^2{x}(1 + 5\sin{x})$
$\displaystyle 40\sin{x} + 8 = 9(1 - \sin^2{x})(1 + 5\sin{x})$
$\displaystyle 40\sin{x} + 8 = 9(1 + 5\sin{x} - \sin^2{x} - 5\sin^3{x})$
$\displaystyle 40\sin{x} + 8 = 9 + 45\sin{x} - 9\sin^2{x} - 45\sin^3{x}$
$\displaystyle 45\sin^3{x} + 9\sin^2{x} - 5\sin{x} - 1 = 0$
you know $\displaystyle \sin{x} = \frac{1}{3}$ 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
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
Hello, Tweety1
$\displaystyle \text{Given: }\:\sin x = \tfrac{1}{3}\,\text{ is a root of the equation: }\:40\sin x + 8 \;=\;9(1+5\sin x)\cos^2\!x$
$\displaystyle \text{Find the other two roots of }\sin x.$
We have: .$\displaystyle 8(5\sin x + 1) \:=\:9(1+5\sin x)\cos^2\!x$
. . . . . . . . $\displaystyle 8(5\sin x + 1) - 9(5\sin x + 1)\cos^2\!x \:=\:0$
. . . . . . . . $\displaystyle (5\sin x + 1)(8 - 9\cos^2\!x) \:=\:0$
$\displaystyle 5\sin x + 1 \:=\;0 \quad\Rightarrow\quad \boxed{\sin x \:=\:\text{-}\tfrac{1}{5}}$
$\displaystyle 8-9\cos^2\!x \:=\:0 \quad\Rightarrow\quad \cos^2\!x \:=\:\tfrac{8}{9} $
$\displaystyle \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}} $
If sin x= 1/3, then $\displaystyle cos^2(x)= 1- sin^2(x)= 1- 1/9= 8/9$ so that we have
$\displaystyle 40sin x+ 8= 9cos^2(x)(1+ 5sin(x))= \frac{40}{3}+ 8= 8(1+ 5/3)$. 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 $\displaystyle 40y+ 8= 9(1- y^2)(1+ 5y)= 9(1+ 5y- y^2- 5y^3)= 9+ 45y- 9y^2- 45y^3$ so that $\displaystyle 45y^3+ 9y^2- 5y- 1= 0$. 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 $\displaystyle 45y^2+ 24y+ 3$ so we can solve the quadratic equation $\displaystyle 45y^2+ 24y+ 3= 0$ 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 $\displaystyle p(x) = x^3 + a_2x^2 + a_1x + a_0$ have a root $\displaystyle r_0$. Let the other roots be $\displaystyle r_1, r_2$.
Since the product of all 3 roots is $\displaystyle -a_0$, and the sum of three roots is $\displaystyle -a_2$, have:
$\displaystyle r_1 + r_2 = -a_2-r_0$ and $\displaystyle r_1 r_2 = \frac{-a_0}{r_0}$ (obviously provided $\displaystyle r_0 \ne 0$).
Thus $\displaystyle r_1, r_2$ are the solutions to: $\displaystyle (y - r_1)(y-r_2) = y^2 - (r_1 + r_2)y + (r_1 r_2) = 0$, so
$\displaystyle r_1, r_2$ are the solutions to: $\displaystyle y^2 - (-a_2-r_0)y + \left(\frac{-a_0}{r_0}\right) = 0$, so
$\displaystyle r_1, r_2$ are the solutions to: $\displaystyle y^2 + (a_2 + r_0)y - \left(\frac{a_0}{r_0}\right) = 0$.
Note that polynomial long division will give you the same result.
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In this case, using the $\displaystyle 45\sin(t)^3 + 9\sin(t)^2 - 5\sin(t) -1 = 0$ others have computed here, you'd get
$\displaystyle \sin(t)^3 + (9/45)\sin(t)^2 - (5/45)\sin(t) - (1/45) = 0$, so $\displaystyle a_2 = 1/5, a_0 = - 1/45, r_0 = 1/3$.
The above becomes: $\displaystyle \sin^2(t) + ((1/5) + (1/3)) \sin(t) - \left(\frac{(-1/45)}{(1/3)}\right) = 0$.
Thus $\displaystyle \sin^2(t) + (8/15) \sin(t) + (1/15) = 0$.
Thus find the solutions to: $\displaystyle 15\sin^2(t) + 8 \sin(t) + 1 = 0$.
(So $\displaystyle (5\sin(t) + 1)(3\sin(t) + 1) = 0$, so $\displaystyle \sin(t) = -1/5$ or $\displaystyle \sin(t) = -1/3$)