How would you solve sinX= - cosX?

TheDude is correct because then you take the arc-tangent of -1 (in tan x=-1) and you get -45. This is correct.

We would use what my professor likes to refer to as the "Aha!" method. We need to know when the sine of something (x) is the same as the negative cosine of that same thing. We know that the only places where |sinx| = |cosx| is when x=(pi)/4, 3(pi)/4, 5(pi)/4, 7(pi)/4, etc., where |sinx|=|cosx|=sqrt(2)/2. So, that narrows it down to four possible values of x. Now, we need to see where the values of sinx and cosx have opposite values. A good trigonometry teacher would have taught you that cosine and sine are both positive in the first quadrant and both negative in the third quadrant. Therefore, their values cannot be opposite for x=(pi)/4 and x=5(pi)/4. That leaves x=3(pi)/4 and x=7(pi)/4. By algebraic inspection (plugging stuff in), we see that both of these values satisfy the equation. Recall that there are an infinite number of solutions; add or subtract any multiple of 2(pi) to your current solutions to find any other solutions.

the laws of nature say you can't divide by cosx. doesn't work in all situations.

you do end up with tanx = -1 either way

which leaves 2 possible solutions for x:

x = 3pi/4

AND

x = 7pi/4

divide each side by cosx so your equation is: tan x = -1