In a right triangle, $\cos (x)^{\circ}=\sin (7 x+9)^{\circ}$. Solve for $x$. Round your answer to the nearest hundredth if necessary.

Step 1 ：Given the equation \(\cos (x)^{\circ}=\sin (7 x+9)^{\circ}\). We know that \(\cos (x) = \sin (90 - x)\) for all x. So, we can rewrite the equation as \(\sin (90 - x) = \sin (7x + 9)\).

Step 2 ：This equation holds true if \(90 - x = 7x + 9\) or \(90 - x = 180 - (7x + 9)\).

Step 3 ：Solving these two equations, we get two potential solutions for x: \(81/8\) and \(27/2\).

Step 4 ：However, we need to check if these solutions are valid in the original equation \(\cos (x)^{\circ}=\sin (7 x+9)^{\circ}\).

Step 5 ：Only the solution \(x = 27/2\) is valid in the original equation \(\cos (x)^{\circ}=\sin (7 x+9)^{\circ}\).

Step 6 ：Final Answer: The solution to the equation \(\cos (x)^{\circ}=\sin (7 x+9)^{\circ}\) is \(\boxed{27/2}\).