Show that $\sin^2 x- 6\sin x-5=0$ has more than one real solution

Question

Without a calculator, prove $\sin^2 x- 6\sin x-5=0$ has more than one real solution.

I have repeatedly solved this but I have only got one solution. Can someone help me out? I want to learn so I can do it next time. Also, this is not homework, I am just studying ahead for next year.

Answer 1

Hint: The polynomial $X^2-6X-5$ has the root $3-\sqrt{14}$ and $3<\sqrt{14}<4$, can you take it from here?

Answer 2

You've figured that any solution $x$ must satisfy $\sin(x)=3\pm\sqrt{14}$. Clearly $3+\sqrt{14}>1$ and we have $-1<3-\sqrt{14}<0$ because $3<\sqrt{14}<4$. Because $\sin(x)$ takes every value in the interval $[-1,1]$, it follows that there exists a real number $x$ such that $\sin(x)=3-\sqrt{14}$. As $\sin(x)$ is a periodic function with period $2\pi$, if $x$ is a solution, then so is $x+2k\pi$ for any integer value of $k$. This shows that there is more than one real solution. In fact, that there are infinitely many.

As far as I know there is no nice way to solve $\sin(x)=3-\sqrt{14}$ for $x$, but this is not the question. All we are asked to do is show that there is more than one solution.