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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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    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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    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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    OH, see. I forgot about that in the OP. I was too fixated on my mistakes. Glad someone remembered.2012-07-17