I can solve the equation in different ways but I'm not getting the right answer when solving as below: $$ \sin x=\cos 2x \\ \cos(\pi/2-x)=\cos 2x \\ \pi/2-x = \pm2x+2\pi k$$
$$x = -\pi /2+2\pi k\\ x = \pi /6-(2\pi k)/3$$
However the second solution should be $x = \pi /6+(2\pi k)/3$. Have checked my math multiple times. Where I'm going wrong?
Remember that $k$ could be either positive or negative, so you could substitute $-k$ instead and you would get the same answer.
$$we\;have\::\\~\\cos(2x)＝sinx\\~\\using:\\~\\sinx＝cos(\textstyle\frac {\pi}{2}－x)\\~\\then\;\;\;the\;\;equation\;\;is\;equivalent\;\;to:\\~\\cos(2x)＝cos(\textstyle\frac {\pi}{2}－x)\;\;\\~\\2x＝∓(\textstyle\frac {\pi}{2}－x)＋2k\pi\;\;\;with\;\;k\;\;is\;integer\\~\\so:\\~\\2x∓x＝∓\textstyle\frac {\pi}{2}＋2k\pi\\~\\2x＋x＝\textstyle\frac {\pi}{2}＋2k\pi\;==>x＝\textstyle\frac {\pi}{6}＋\textstyle\frac {2k\pi}{3}\\~\\~2x－x\;＝－\textstyle\frac {\pi}{2}＋2k\pi\;==>\;\;x＝－\textstyle\frac {\pi}{2}＋2k\pi$$
