Re: answerer's question at Yahoo! Answers regarding trigonometric equation
Hello answerer,
We are given to solve:
$$3\sin^2(x)-2\sin(x)-3=0$$ where $$0<x<2\pi$$
Recognizing that we have a quadratic in $\sin(x)$, we may use the quadratic formula to state:
$$\sin(x)=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(-3)}}{2(3)}=\frac{2\pm\sqrt{40}}{6}=\frac{1\pm\sqrt{10}}{3}$$
Since we require $$-1\le\sin(x)\le1$$ we discard the positive root, and we are left with:
$$\sin(x)=\frac{1-\sqrt{10}}{3}$$
Hence:
$$x=\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)$$
Since this is less than zero, we need to add $$2\pi$$ to get the equivalent angle in the required interval:
$$x=2\pi+\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)$$
Now, this is the 4th quadrant solution, but we should observe there is also a 3rd quadrant solution, given by:
$$x=\pi-\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)$$
Note: this comes from the identity $$\sin(\pi-\theta)=\sin(\theta)$$.
If we are to use a calculator to obtain decimal approximations, then:
$$x=2\pi+\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)\approx5.47828834852818$$
$$x=\pi-\sin^{-1}\left(\frac{1-\sqrt{10}}{3} \right)\approx3.9464896122412$$
To answerer and any other guests viewing this topic, I invite and encourage you to post other trigonometry questions here in our http://www.mathhelpboards.com/f12/ forum.
Best Regards,
Mark.