$-\sin(x) = \cos(x)$

$\dfrac{\sin(x)}{\cos(x)} = -1$

$\tan(x) = -1$

solve that for $x$

While this method is correct, I tend to be wary of solving equations through division, as you never know if you are dividing by 0 and thus losing solutions. In this case I would probably lean more towards making use of a compound angle identity $\displaystyle \begin{align*} \sin{ \left( \alpha + \beta \right) } \equiv \sin{ \left( \alpha \right) } \cos{ \left( \beta \right) } + \cos{ \left( \alpha \right) } \sin{ \left( \beta \right) } \end{align*}$.

So with this equation we have

$\displaystyle \begin{align*} -\sin{ \left( x \right) } &= \cos{ \left( x \right) } \\ 0 &= \sin{ \left( x \right) } + \cos{ \left( x \right) } \\ 0 \cdot k &= \sin{ \left( x \right) } \cdot k + \cos{ \left( x \right) } \cdot k \\ 0 &= \sin{ \left( x \right) } \cos{ \left( y \right) } + \cos{ \left( x \right) } \sin{ \left( y \right) } \end{align*}$

and here $\displaystyle \begin{align*} \cos{ \left( y \right) } = \sin{ \left( y \right) } = k \end{align*}$.

As $\displaystyle \begin{align*} \cos{ \left( y \right) } = \sin{ \left( y \right) } \end{align*}$ the only possible value of y in the first cycle is $\displaystyle \begin{align*} \frac{\pi}{4} \end{align*}$, where both the sine and cosine are $\displaystyle \begin{align*} k = \frac{1}{\sqrt{2}} \end{align*}$, thus

$\displaystyle \begin{align*} 0 &= \sin{ \left( x \right) } \cos{ \left( \frac{\pi}{4} \right) } + \cos{ \left( x \right) } \sin{ \left( \frac{\pi}{4} \right) } \\ 0 &= \sin{ \left( x + \frac{\pi}{4} \right) } \\ n\,\pi &= x + \frac{\pi}{4} \textrm{ where } n \in \mathbf{Z} \\ x &= n\,\pi - \frac{\pi}{4} \\ x &= \frac{\left( 4\,n - 1 \right) \, \pi}{4} \end{align*}$