Imagine \(\angle A\) at the top of the triangle and \(\angle B\) at the bottom. (or used the attached picture for reference)
\(\sin A= \frac{opposite}{hypotenuse}\)
\(\cos B = \frac{adjacent}{hypotenuse}\)
One thing to note, however, is that the side opposite \(\angle A\) is the same side as the one adjacent to \(\angle B\)! Thus \(\sin A=\cos B\).
Substitute these values for \(2x+0.1\) and \(4x-0.7\) and solve.
\(4x-0.7=2x+0.1 \\ 4x=2x+0.8 \\ 2x=0.8\\ \boxed{x=0.4}\)
