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This is the problem I have in my trigonometry packet. Can anyone give it a try?

I factored by doing:

$$\sin x(3\sin x - 2) = 1$$

Then divided $\sin x$:

$$ 3\sin x - 2 = \frac{1}{\sin x}$$

But then I got stuck.

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I'll post the edit showing what I've tried. :) – David Apr 6 '12 at 15:50
The edit is up! – David Apr 6 '12 at 15:55
Once again: $360^\circ$ looks different from $360^o$ and the first is the standard notation. – Michael Hardy Apr 6 '12 at 22:30
up vote 3 down vote accepted

We can rewrite the equation as $$3(\sin x)^2-2\sin x-1=0.$$ Temporarily, let $w=\sin x$. So our equation can be rewritten as $$3w^2-2w-1=0.$$ This is a quadratic equation, so the solutions can be found by using the Quadratic Formula. But it so happens that our polynomial factors nicely: $$3w^2-2w-1=(3w+1)(w-1).$$ So the roots of the polynomial are given by $w=-1/3$ and $w=1$.

We want to find all angles between $270$ degrees and $360$ degrees whose sine is either $-1/3$ or $1$. There is no angle in the fourth quadrant whose sine is $1$. But there is one whose sine is $-1/3$.

The first quadrant angle whose sine is $1/3$ is (calculator) about $19.47$ degrees. So the fourth quadrant angle whose sine is $-1/3$ is about $360-19.47$ degrees.

Remark: If you enter the number $-1/3$ in the calculator, and press the "$\sin^{-1}$" button, then, if your calculator is in degree mode, you will get something like $-19.471221$. The negative sign means that we are travelling clockwise from the positive $x$-axis. To get the answer in the more conventional style, between $270$ and $360$, add $360$ to the calculator's (negative) answer.

The type of factoring that you did is in general not a useful way of solving quadratic equations. In the notation I used above, we have $w(3w-2)=1$. Transforming to the standard shape quadratic equation $3w^-2w-1=0$ works nicely. The strategies you were trying unfortunately complicate the problem instead of simplifying it. It is important for future problems to be able to recognize "disguised" quadratics.

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Thanks. We were doing problems (non-quadratic ones) where we needed to factor and so I used it on this one, not knowing it was a quadratic equation "in disguise". :) – David Apr 6 '12 at 16:18
@David: Mostly when you need to factor, it is useful to have all the terms together. There are some exceptions, like $x^2(x-3)=2(x-3)$, where it may be useful to note that $x$ can be $3$, but if it isn't we can cancel and get $x^2=2$. But even here, bringing stuff together works nicely and is not much more work, for $x^2(x-3)-2(x-3)=(x-3)(x^2-2)$ and you conclude that $x=3$ or $x=\pm\sqrt{2}$. – André Nicolas Apr 6 '12 at 16:23
Is there any way I can direct message you? I have some other problems I think you can help me with. :] – David Apr 6 '12 at 16:51
@David: I think it is best to ask questions on this site. As you can see, answers ordinarily come very quickly. Often, there will be a variety of approaches. – André Nicolas Apr 6 '12 at 17:07



$$\sin(x)=1 \, {\rm or} \sin(x)=-\frac{1}{3} \,.$$

$\sin(x)=1$ has no root in that interval, while $\sin(x)=-\frac{1}{3}$ can be found with a calculator...

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Hint: let $A = \sin (x)$ and solve the resulting polynomial.

One of the solutions will be $90$ degrees (or $\pi /2$ radians), which is not in the interval under consideration.

Edit - to comment on your recent edit.

The most common reason to factor something (at this level) is to use the "zero product property" (Integral Domain) when one side of the equation is ZERO!

So to start factoring before putting the equation in standard form is usually going to result in a dead end. Hopefully you can see the reasoning for a simpler example like $$x^2 - x = 6$$

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