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I am trying to understand how to solve the equation

$$2\sin^2x + 3\sin x + 1 = 0.$$

Please give hints.

Bilbo
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    Hi, welcome to math.SE! Can you show what you've done so far, and what you're having trouble with? We prefer to help you understand and solve things on your own, rather than being a homework-solving service. – Jonathan Christensen Dec 20 '12 at 23:39
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    Please avoid using "ASAP" when asking for help here. It's impolite. – Asaf Karagila Dec 20 '12 at 23:39
  • @Bilbo: I edited your question, changing the tone slightly. In the future, please do not demand a solution "ASAP" to the problem you are working, that is not the purpose of Math Stack Exchange. We will help you understand the problem, but we are not here to do your homework for you. – Eric Naslund Dec 20 '12 at 23:42
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    Nevermind, I just had to factor it into (sinx + 1)(2sinx + 1). – Bilbo Dec 20 '12 at 23:42
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    Have you thought of substituting $y=\sin x$? – Mark Bennet Dec 20 '12 at 23:43
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    @EricNaslund The edited version of this question does not only change the tone slightly, rather it contradicts explicit demands made by the OP in the original version, namely, to solve this for them and to show the steps, ASAP. One might view such a modification as problematic. – Did Dec 20 '12 at 23:49
  • @did: You make a good point, I almost entirely changed the question. It is debatable whether or not this was the correct thing to do, but let me briefly explain why I made this decision. It was Bilbo's first question this site, and it was written in an inappropriate way. First time users often do not know how to ask questions properly here, and I felt it was best to improve the question and leave note explaining to Bilbo not to do this in the future. This removes the low quality content, and hopefully teaches a new user in a positive way how to use the site correctly. – Eric Naslund Dec 20 '12 at 23:58
  • @EricNaslund You surely guessed that I had guessed the reasons you explain for doing so. Your motivation is laudable but whether the method is effective or not remains to be seen. Note that this is once again doing the OP's job for them. Note also that a note explaining to the OP what to do in the future was already posted as a comment. – Did Dec 21 '12 at 00:06
  • @did: I think it was effective in this case, look at the OP's newest question: http://math.stackexchange.com/questions/262961/help-solving-trigonometry-equation I do believe in helping the OP out as much as possible if they are a new user, even if that means "doing [their] job for them." From my experience, most users want to use the site properly, and just need some guidance at the start. It's the few bad apples that stand out... – Eric Naslund Dec 21 '12 at 00:07
  • @EricNaslund You call this effective? Still no effort by the OP and, quite probably, somebody collecting the answers to the successive items of their homework. – Did Dec 21 '12 at 00:10
  • @did: He was asking for a hint, which seems fair enough. At the very least, he learned something about the tone in which he should ask the question. There may not be that much effort present, but the new question undeniably looks significantly better. His comment on my answer, "I always forget to use factoring as an option to solve when solving trigonometry equations, thank you." is also reassuring. Here is my question: Do you think it would have been better if I had originally left this question untouched so that it was downvoted into oblivion? Would his next question have improved? – Eric Naslund Dec 21 '12 at 00:17
  • @EricNaslund Sorry but He was asking for a hint is not accurate. The OP's version: Could someone solve this for me? Please show your steps. Your version: Please give hints. About the question in your comment: the two options you describe are not the only ones. A third one (hopefully quite frequent) is the OP reacting to the constructive comments made at them (such as the one Jonathan posted) and reaching by themselves an acceptable version of their question. (On a more personal note, if you assigned to yourself the task to make acceptable every new MSE post, you are in for a treat...) – Did Dec 21 '12 at 00:26
  • @did I think you misunderstood me, his newest question was in fact asking for a hint. (Not this question) Which is why I said at the very least he learned something about the tone in which he should ask. – Eric Naslund Dec 21 '12 at 00:29
  • let us continue this discussion in chat – Eric Naslund Dec 21 '12 at 00:30
  • @EricNaslund Ach so. I did misunderstand, sorry about that. – Did Dec 21 '12 at 00:30

2 Answers2

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Hint: Let $y=\sin x$. Solve for $y$ first, then solve for $x$.

Jonathan Christensen
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Just treat $\sin x$ as a variable, and solve the polynomial.

Eric Naslund
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Ken
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