Answer:
If [tex]sin x = 0[/tex] we get [tex]2 (0) (\pm1) = 0[/tex] which is true. so we have a first set of solutions given by
[tex]x= 0; x = \pi[/tex];
Else, if [tex]sin x \ne 0[/tex] we can divide both sides by it and we get
[tex]2 cos x = \frac 1 {cos x} \rightarrow 2cos^2x =1 \rightarrow cos^2x =\frac12\\cos x= \pm \frac{\sqrt2}2[/tex]
Which gives us a second set of solutions given by [tex]x= \frac{\pi}4;x= \frac{3\pi}4; x= \frac{5\pi}4;x= \frac{7\pi}4[/tex];
We can group all solutions (doesn't matter, but it's more elegant!) by writing
[tex]x= k\frac{\pi}4; k\in \{0,1,3,4,5,7\}[/tex]
