katex fix

sin x = 2.md

@@ -19,22 +19,40 @@ $$
 u - u^{-1} = 4i
 $$
 $$
-u^2 - 1 = 4iu \\ $$$$
+u^2 - 1 = 4iu \\ $$
+$$
 u^2 - 4iu - 1 = 0
-$$$$
-u^2 - 4iu - 4 = -3 $$$$
-(u - 2i)^2 = -3 \\ $$$$
-u - 2i = \pm \sqrt {-3} $$$$
-u = 2i \pm \sqrt {-3} \\ $$$$
+$$
+$$
+u^2 - 4iu - 4 = -3
+$$
+$$
+(u - 2i)^2 = -3 \\
+$$
+$$
+u - 2i = \pm \sqrt {-3} 
+$$
+$$
+u = 2i \pm \sqrt {-3} \\
+$$
+$$
 u = i(2 \pm \sqrt 3)
 $$
 
 Substitute back into $u$, for $n \in \mathbb{Z}$:
 
 $$
-e^{ix} = i(2 \pm \sqrt 3) \\ $$$$
-ix = \ln (i(2 \pm \sqrt 3)) \\ $$$$
-ix = \ln i + 2\pi n+ \ln(2 \pm \sqrt 3) \\ $$$$
-ix = \frac {i\pi} 2 + 2\pi n + \ln(2 \pm \sqrt 3) $$$$
+e^{ix} = i(2 \pm \sqrt 3) \\ 
+$$
+$$
+ix = \ln (i(2 \pm \sqrt 3)) \\ 
+$$
+$$
+ix = \ln i + 2\pi n+ \ln(2 \pm \sqrt 3) \\
+$$
+$$
+ix = \frac {i\pi} 2 + 2\pi n + \ln(2 \pm \sqrt 3)
+$$
+$$
 x = \frac \pi 2 - i\ln(2 \pm \sqrt 3) + 2\pi n
 $$