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Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$ $$\cos^2x-\sin^2x=\sin$$

Edit $$1-\sin^2x-\sin^2x=\sin x$$ $$2\sin^2 x+\sin x-1=0$$ $\sin x=a$ $$2a^2+a-1=0$$ $$(a+1)(2a-1)=0$$ $$x=-1,\dfrac{1}{2}$$ $$x=\sin^{-1}(.5)=30^{\circ}=\dfrac{\pi}{6}$$ $$x=sin^{-1}(-1)=-90^{\circ}=-\dfrac{\pi}{2}$$

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    What are you doing from the 2nd to the 3rd line? It looks like you're trying to add $sin x$ even though it is within a product in parantheses. I'm pretty sure you know that you can't do that.2012-07-16
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    I was just trying to get something accomplished on here so I can get useful help. What do I do after the second line?2012-07-16

2 Answers 2

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What you have is:

$$2\sin^2 x+\sin x-1=0$$

And let $\sin x = a$ so you'll have to solve a quadratic equation for $a$.

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    @AustinBroussard: Sorry, I don't understand. you just substitute $\sin x$ with something more familiar and less annoying! If that's what you meant. When you find the values of $a$ you'll have values of $\sin x$ accordingly.2012-07-17
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    @AustinBroussard: Suppose $\sin x =a$, and solve the equation $a^2+a-1=0$. What values of $a$ you can find?2012-07-17
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    @AustinBroussard: Yes, exactly. But your answer is not correct, how you got $a=\frac{\sqrt 2}2$?2012-07-17
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    @AustinBroussard: You got: $(a+1)(2a-1)=0$, right? What are the values of $a$? Please first make sure you understand what I am saying and you have the correct answer, then you can accept my answer!2012-07-17
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    $a=-1$ and $\dfrac{1}{2}$. Can you delete your older comments because the site hates an overload of comments.2012-07-17
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    @AustinBroussard: Great, that's it. Now you have $\sin x=-1$ and $\sin x= \frac 12$ while $-\pi , can you find the values of $x$ from there?2012-07-17
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    Lets see. $$x=\sin^{-1}(.5)=30^{\circ}=\dfrac{\pi}{6}$$ $$x=sin^{-1}(-1)=-90^{\circ}=-\dfrac{\pi}{2}$$2012-07-17
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    @AustinBroussard: That's correct, well done. And I see no need to delete those comments while they show your effort to solve the problem.2012-07-17
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    Thanks a lot for the help and cooperation. I ask a lot of questions to get the full understanding and so I can use it in my future problems. Thanks a lot for your time and brain!2012-07-17
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The calculation is almost completely correct. You reached the two possibilities $\sin x=\frac{1}{2}$ and $\sin x=-1$.

We are interested in solutions in the interval $-\pi \lt x\le \pi$.

Certainly $x=\frac{\pi}{6}$ is a solution, since $\sin(\pi/6)=\frac{1}{2}$. But there is another $x$ in our interval whose sine is $\frac{1}{2}$, namely $x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$. A look at the graph of $y=\sin x$ shows this. You can do a partial verification by calculator, by asking it to compute $\sin(5\pi/6)$, the sine of $150^\circ$.

There is only one place $x$ in our interval where $\sin x=-1$, so that part is fully correct.

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    So the two answers I have are incorrect? $\dfrac{\pi}{6}$ and $-\dfrac{\pi}{2}$2012-07-17
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    The two answers are correct, I wrote that they are correct. You missed one answer, $\frac{5\pi}{6}$. So there are three answers, of which you got two.2012-07-17
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    $30^\circ -180^\circ =150^\circ$? Is that why there is a third answer?2012-07-17
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    You have it sort of backwards, $180^\circ-30^\circ=150^\circ$, which is why $150^\circ$ is a solution.2012-07-17
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    But why is it so important to find an answer in the second quadrant? I'm just confused as to why you have to have the $150^\circ$2012-07-17
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    @AustinBroussard: Because the question asks for **all** solutions $x$ with $-\pi \lt x \le \pi$. The number $\frac{5\pi}{6}$ is in the interval asked about.2012-07-17
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    OH, see. I forgot about that in the OP. I was too fixated on my mistakes. Glad someone remembered.2012-07-17