From Förberedande kurs i matematik 1

Apart from the fact that there is a \displaystyle 5x, this is a normal trigonometric equation of the type \displaystyle \sin y = a\,. If we are only interested in solutions which satisfy \displaystyle 0\le 5x\le 2\pi, then a sketch of the unit circle shows that there are two such solutions, \displaystyle 5x = \pi/4 and the reflectionally symmetric solution \displaystyle 5x = \pi - \pi/4 = 3\pi/4\,.

All of the equation's solutions are obtained from all values of \displaystyle 5x which differ by a multiple of \displaystyle 2\pi from either of these two solutions,

\displaystyle 5x = \frac{\pi}{4} + 2n\pi\qquad\text{and}\qquad 5x = \frac{3\pi}{4} + 2n\pi\,,

where n is an arbitrary integer.

If we divide both of these by 5, we obtain the solutions expressed in terms of x alone,

\displaystyle x = \frac{\pi}{20} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{3\pi}{20} + \frac{2}{5}n\pi\,,

where n is an arbitrary integer.