The question says "find all real numbers in the interval $[0,2\pi)$ that satisfy $$3\sin(x)^2-\sin(x)=0$$ The answers it gives are $\{0, 0.3, 2.8, \pi \}.$ I rearranged the equation and used the quadratic formula, and figured out the 0 and PI, but I'm not sure how the got the decimals.

  • $\begingroup$ why would you use quadratic formula to find the roots of $0=3\sin^2 x - \sin x=\sin x(3\sin x - 1)$ \to $\sin x = 0, \sin x = \frac 13$ $\endgroup$ – abel Apr 28 '15 at 23:04
  • $\begingroup$ Because an example problem we did rearranged the equation and factored it to get the answer, and using the quad. formula in place of the factoring also got to the answers. It was the only way I knew to try and go about it. $\endgroup$ – windy401 Apr 28 '15 at 23:09
  • $\begingroup$ see my answer below. $\endgroup$ – abel Apr 28 '15 at 23:11

why would you use quadratic formula to find the roots of $$0=3\sin^2 x - \sin x=\sin x(3\sin x - 1) \to \sin x = 0, \sin x = \frac 13$$

the decimal come from the equation $$\sin x = \frac 13 \to x = \sin^{-1}(1/3), \pi-\sin^{-1}(1/3).$$

  • $\begingroup$ Ok, I need to learn more about factoring. I know the basic stuff but how you get to that, I don't see. The rest I get though. $\endgroup$ – windy401 Apr 28 '15 at 23:43
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    $\begingroup$ @windy401, it is like factoring $y^2 - 3y = 0.$ you can factor out a $y$ from the terms $3y$ and $y^2.$ it gives you $0=y^2 - 3y = y(3y-1)$ so that you have either $y = 0$ or $3y - 1= 0.$ in your equation, $\sin x$ plays the role of $y.$ $\endgroup$ – abel Apr 28 '15 at 23:49
  • $\begingroup$ Ok, I see where the $1/3$ comes from now... But the $y2−3y=y(3y−1)$, I'm just not seeing how it's converted from one to the other. $\endgroup$ – windy401 Apr 29 '15 at 0:02
  • $\begingroup$ if you have a quadratic $ax^2 + bx + c = 0$ and ig either $b$ or $c$ is missing, then it is easier to factor than using the quadratic formula. you reserve the formula for the case when all coefficients are present. you are making good progress. $\endgroup$ – abel Apr 29 '15 at 0:05
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    $\begingroup$ @windy401 Lets take the fact that $y^{2} = y * y$ this means that $y^{2} - 3y = y * y - 3y$ Now this type of factoring were doing is called "COMMON factoring" So we need to take out what is "common" between the two terms and as we see both terms have a $y$ which means we can "common factor" a $y$ and then when we take something out we need to divide both terms by what we took out so $y(\frac{3y^{2}}{y} - \frac{y}{y}) = y(3y-1)$ $\endgroup$ – JackV Apr 29 '15 at 0:14

Well first off as abel said you dont need the quadratic formula you can just factor out a $sinx$ to get $sinx(3sinx - 1) = 0$ So we can easily see $sinx = 0$ and $sinx = 1/3$ so first lets explore

$$sinx = 0$$ so we take $sin^{-1}x$ of both sides to get

$$x = sin^{-1}0$$

which gives us $x = 0$ but we also know $sinx = 0$ at $\pi$ so thats how we get the answers $0, \pi$

Now we have

$$sinx = 1/3$$

$$x = sin^{-1}(\frac{1}{3})$$

$$x = 0.3$$

But if we look at the unit circle we can see that there are obviously two points in which $sinx = 1/3$ so we need to subtract our current answer from pi to get our next answer so

$$x = \pi - 0.3$$

$$x = 2.8$$

  • $\begingroup$ $\sin^{-1}(1/3) $ is not exactly 1/3 $\endgroup$ – Narasimham Apr 29 '15 at 18:52
  • $\begingroup$ @Narasimham I am aware of that but the person already told us the answers they were supposed to find and these were it but you are correct in the fact that i could have been a lot more accurate in my answer $\endgroup$ – JackV Apr 30 '15 at 16:07

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