Can someone help me on this?

Answer:
Step-by-step explanation:
We know that cosx = sin([tex]\pi[/tex] - x)
We can apply it here:
sin(55°) = cosx
cos([tex]\pi[/tex] - 55°) = cosx
Now we apply the arccos to both sides:
x = [tex]\pi[/tex] - 55° + 2k[tex]\pi[/tex] or x = -[tex]\pi[/tex] - 55° + 2k[tex]\pi[/tex]