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I was helping someone with her homework the other day, doing trigonometry problems. I ran into something which I wasn't too sure how to work out.

The question was:

Find every possible answer in terms of $\pi$. $$(\sin^2 x)+(\sin x)-2=0$$

I broke it down like this (but I am not sure if it's right).

\begin{align}\sin(\sin(x))+\sin x -2&=0\\ \sin(2\sin x)-2&=0\\ 2\sin^2(x)-2&=0\\ 2\sin^2(x)&=2\\ \sin^2(x)&=2/2\\ \sin^2(x)&=1\\ \sin(x)&=1\end{align}

From there, she could find the answer in terms of $\pi$ on her own, but I'm not sure if this was right at all.

Peter Mortensen
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alex
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    It's a quadratic in the variable $\sin x$. – David Mitra May 24 '17 at 12:54
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    It's been mentioned that you interpreted sin^2(x) incorrectly. I'd like to point out that even if your interpretation was correct the next step isn't justified. sin(sin(x)) + sin x wouldn't end up being sin(2sinx) and then even if that was true you can't just pull out a constant multiple from the sine. – Dason May 24 '17 at 15:44

4 Answers4

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You appear to have done $$\begin{align}\sin^2x+\sin x-2=0\\\sin(\sin x)+\sin x-2=0\\\sin(2\sin x)-2=0\\2\sin(\sin x)-2=0\\\sin(\sin x)=1\\\sin^2x=1\\\sin x=1\end{align}$$

It seems you are confused by the meaning of $\sin^2x$. It does not mean $\sin(\sin(x))$, it means $(\sin(x))^2$. So if you let $u=\sin x$, then you have $u^2+u-2=0$, which is an easy quadratic to solve.

This gives solutions of $u=\sin x=1,-2$. But for real $x$, $\sin x\neq -2$ so you just need to solve the equation $\sin x=1\implies x=\frac{\pi}{2}+n\pi$ for $n\in\Bbb Z$.

John Doe
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You could try to substitute and see this is a polynomial question, or else directly:

$$0=\sin^2x+\sin x-2=(\sin x+2)(\sin x-1)\implies \sin x=\begin{cases}-2\\{}\\\text{ or}\\{}\\\;\;1\end{cases}$$

The first option is impossible, so it must be $\;\sin x=1\;$ and etc.

DonAntonio
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  • Note that $\sin x = 2$ does have infinitely many solutions, just not any real solutions. – MPW May 24 '17 at 18:12
  • @MPW Someone asking such a basic, elementary question is very doubtable he even knows complex numbers, let alone complex analysis... – DonAntonio May 24 '17 at 18:15
  • No doubt. Just saying. The nonreal solutions are the values $$(4k+1)\frac{\pi}{2} \pm i\log(2+\sqrt{3})$$ for integral $k$, if I've computed correctly. – MPW May 24 '17 at 18:19
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$$\sin^2 x +\sin x-2=0$$

therefore

$$\sin^2 x+\sin x=2$$

But $\sin x$ varies between $-1$ and $1$, and therefore $\sin^2 x$ between $0$ and $1$. Thus, for their sum to be $2$, they must simultaneously equal $1$.

But this happens when and only when $x$ is $\frac{\pi}{2}$ more than an even multiple of $\pi$: and this is the answer.

$$x = 2k + \frac{\pi}{2}, k \in \mathbb{Z}$$

mtheorylord
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Matt Gutting
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Hint : $\sin^2x+\sin x-2=(\sin x-1)(\sin x+2)$

user362325
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