MHB Answerer's question at Yahoo Answers regarding a trigonometric equation

AI Thread Summary
The discussion addresses a trigonometric equation, specifically 3sin²x - 2sinx - 3 = 0, within the interval 0 < x < 2π. The equation is identified as a quadratic in sin(x), and the quadratic formula is applied to find the roots. After determining the valid solution, it is noted that the sine value leads to two angles: one in the fourth quadrant and another in the third quadrant. Decimal approximations for these angles are provided for clarity. The conversation encourages further trigonometry questions to foster engagement in the forum.
MarkFL
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Here is the question:

PRE-CALC QUESTION!? HELP !?

0<x<2π
3sin^2x - 2sinx - 3 = 0

You can use a calculator and there's only 2 answers

Here is a link to the question:

PRE-CALC QUESTION!? HELP !? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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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.
 
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