# Trig Equation

• Dec 28th 2012, 02:36 PM
kinhew93
Trig Equation
1 + 4sin[A](3sin[A] - 4) = cosec[A](cosec[A]-4)

Find the possible values of sin[A]

I tried writing as a quartic equation but I couldn't factorise it and I am now totally void of ideas. Could someone point me in the right direction?
• Dec 28th 2012, 02:52 PM
Plato
Re: Trig Equation
Quote:

Originally Posted by kinhew93
1 + 4sin[A](3sin[A] - 4) = cosec[A](cosec[A]-4)
Find the possible values of sin[A]

I have no idea if this helps, but I would try:

$\csc(A)(\csc(A)-4)=\frac{1}{\sin^2(A)}-\frac{4}{\sin(A)}$
• Dec 28th 2012, 04:57 PM
Soroban
Re: Trig Equation
Hello, kinhew93!

Quote:

$\text{Find the possible values of }\sin\theta:$

. . . $1 + 4\sin\theta (3\sin\theta - 4) \:=\: \csc\theta(\csc\theta -4)$

We have: . . . . . $1 + 4\sin\theta(3\sin\theta - 4) \;=\;\frac{1}{\sin\theta}\left(\frac{1}{\sin\theta } - 4\right)$

n . . . . . . . . . . . $1 + 12\sin^2\!\theta - 16\sin\theta \;=\;\frac{1-4\sin\theta}{\sin^2\!\theta}$

n . . . . . . . . $\sin^2\!\theta + 12\sin^4\!\theta - 16\sin^3\!\theta \;=\;1 - 4\sin\theta$

$12\sin^4\theta - 16\sin^3\theta + \sin^2\theta + 4\sin\theta - 1 \;=\;0$

Factor: . $(\sin\theta-1)(2\sin\theta-1)(2\sin\theta+1)(3\sin\thetau-1) \;=\;0$

Therefore: . $\sin\theta \;=\;\begin{Bmatrix} 1 \\ \pm\frac{1}{2} \\ \frac{1}{3} \end{Bmatrix}$