Problem relating to trigonometry 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.
 A: 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 $n\in\Bbb Z$.
A: 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.
A: $$\sin^2 x +\sin x-2=0$$
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.
$$x = 2k + \frac{\pi}{2}, k \in \mathbb{Z}$$
A: Hint : $\sin^2x+\sin x-2=(\sin x-1)(\sin x+2)$
