Given,
[tex]\sin (7x+1)=\cos (x+33)[/tex]We know that,
[tex]\sin \theta=\cos (90-\theta)[/tex][tex]\begin{gathered} \cos (90-(7x+1))=\cos (x+33) \\ \text{equating the parameters of cosines on both sides we get,} \\ 90-7x-1=x+33 \\ 89=7x+x+33 \\ 8x=89-33 \\ 8x=56 \\ x=\frac{56}{8} \\ x=7 \end{gathered}[/tex][tex]\begin{gathered} \text{Angle 1 =(7x+1)} \\ \text{Angle 1 =(7}\times7)\text{+1)} \\ \text{Angle 1 =}49+1 \\ \text{Angle 1 =5}0^{\circ} \end{gathered}[/tex][tex]\begin{gathered} \text{Angle 2 =(}x+33\text{)} \\ \text{Angle 2 =(}7+33\text{)} \\ \text{Angle 2 = 40}^{\circ} \end{gathered}[/tex]