My progress using this site to solving trigonometry equation about reflection, shift and periodicity

Question

I joined with this site a day ago. Because i have a task about trigonometry and an exam in next 2 weeks. From zero knowledge about trigonometry and now i just have a spirit to learn trigonometry.

This is my first progress to practice.

Find the solution of x between $0^\circ$ and $360^\circ$ of this equation : $$\sin(x+60^\circ)=\cos(x-20^\circ)$$

Answer 1

Converting left hand side using $\cos(1×90^\circ-x)=\sin(x)$ $$\sin(x+60^\circ)=\cos(90^\circ-(x+60^\circ))=\cos(30^\circ-x)$$ Substitution into the problem again, and i get : $$\cos(30^\circ-x)=\cos(x-20^\circ)$$ Using arccosine, then i get : $$30^\circ-x=x-20^\circ$$ $$30^\circ+20^\circ=x+x$$ $$2x=50^\circ$$ $$x=25^\circ$$ Converting left hand side using $\cos(5×90^\circ-x)=\sin(x)$ $$\sin(x+60^\circ)=\cos(450^\circ-(x+60^\circ))=\cos(390^\circ-x)$$ Substitution into the problem again, i get : $$\cos(390^\circ-x)=\cos(x-20^\circ)$$ Using arccosine, then i get : $$390^\circ-x=x-20^\circ$$ $$2x=410^\circ$$ $$x=205^\circ$$ Converting left hand side using $\cos(9×90^\circ-x)=\sin(x)$ $$\sin(x+60^\circ)=\cos(810^\circ-(x+60^\circ))=\cos(750^\circ-x)$$ Substitution into the problem again, i get : $$\cos(750^\circ-x)=\cos(x-20^\circ)$$ Using arccosine, then i get : $$750^\circ-x=x-20^\circ$$ $$2x=770^\circ$$ $$x=385^\circ$$ Using periodicity of cosine $$x=385^\circ-360^\circ$$ $$x=25^\circ$$ Because the solution is repeating, then i should stop. Finally i knew the set of solution are : $$[25^\circ,205^\circ]$$ What i get so far, using math stack exchange are :

1. $\sin(x)=\cos(1×90^\circ-x)$ this called cofunction relation.
2. $\sin(x)=\cos(5×90^\circ-x)=\cos(9×90^\circ-x)=\cos(13×90^\circ-x)$ the multiplier always leap four, because in unit circle there are four quadrant.
3. From point 2, i really realize to adding $+k×360^\circ$ after i converting sine into cosine, to fastening my calculation process.
4. I knew about sum angle formula, arccosine and unit circle always helpimg in trigonometry problem.

Thank you so much to all users that helped me reached in this trigonometry knowledge. Thanks for your patience, time, information, suggestion, opinion and solution to my problem. God bless you.