2(sin^-1x)cubed+5(sin^-1x)squared-1=0
Helppp solve the equation. Please?
As well as find the area and perimeter of the triangle.
cos(a) equals approximately -.125, b=8, c=8
Helpppp!
-Thanks
Let $\displaystyle y = sin^{-1}(x)$
Then we need to solve
$\displaystyle 2y^3 + 5y - 1 = 0$
The rational root theorem says that if there is a rational root for this polynomial equation it is of the form $\displaystyle y = \pm 1, \pm \frac{1}{2}$.
Trial and error shows that y = -1/2 is indeed a solution. This means we can draw out a factor of the form 2y + 1:
$\displaystyle 2y^3 + 5y - 1 = (2y + 1)(y^2 + 2y - 1)$
by any method of polynomial division you choose.
So finishing the solution I get that
$\displaystyle y = -\frac{1}{2}, 1 \pm \sqrt{2}$
So we need to solve for the three solutions:
$\displaystyle x = sin^{-1} \left ( -\frac{1}{2} \right )$
and
$\displaystyle x = sin^{-1}(1 + \sqrt{2} )$
and
$\displaystyle x = sin^{-1}(1 - \sqrt{2} )$
The last two are impossible, as $\displaystyle |1 \pm \sqrt{2}| > 1$. So we only have
$\displaystyle x = sin^{-1} \left ( -\frac{1}{2} \right ) \implies x = -\frac{\pi}{6}~rad$
-Dan
To get the third side of the triangle, use the Law of Cosines. Then you can get the perimeter easily.
To get the area you can use Heron's formula.
-Dan