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

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

Austin Broussard
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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. – us2012 Jul 16 '12 at 23:53
  • I was just trying to get something accomplished on here so I can get useful help. What do I do after the second line? – Austin Broussard Jul 16 '12 at 23:54

2 Answers2

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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 $a$.

Gigili
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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. – Gigili Jul 17 '12 at 00:09
  • @AustinBroussard: Suppose $\sin x =a$, and solve the equation $a^2+a-1=0$. What values of $a$ you can find? – Gigili Jul 17 '12 at 00:15
  • @AustinBroussard: Yes, exactly. But your answer is not correct, how you got $a=\frac{\sqrt 2}2$? – Gigili Jul 17 '12 at 00:19
  • @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! – Gigili Jul 17 '12 at 00:22
  • $a=-1$ and $\dfrac{1}{2}$. Can you delete your older comments because the site hates an overload of comments. – Austin Broussard Jul 17 '12 at 00:23
  • @AustinBroussard: Great, that's it. Now you have $\sin x=-1$ and $\sin x= \frac 12$ while $-\pi <x\leq \pi$, can you find the values of $x$ from there? – Gigili Jul 17 '12 at 00:25
  • Lets see. $$x=\sin^{-1}(.5)=30^{\circ}=\dfrac{\pi}{6}$$ $$x=sin^{-1}(-1)=-90^{\circ}=-\dfrac{\pi}{2}$$ – Austin Broussard Jul 17 '12 at 00:29
  • @AustinBroussard: That's correct, well done. And I see no need to delete those comments while they show your effort to solve the problem. – Gigili Jul 17 '12 at 00:35
  • 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! – Austin Broussard Jul 17 '12 at 00:36
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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.

André Nicolas
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