A force F varies with distance x and time t according to the...

By the principle of homogeneity, the dimensions of each term added together must be the same as the dimensions of the LHS.

Therefore, $[A] = [F]$ and $[C] = [F]$.
This implies that the ratio $\dfrac{A}{C}$ is dimensionless, i.e., \left[\dfrac{A}{C}\right] = [M^0 L^0 T^0]$.<br><br>The arguments of trigonometric functions must be dimensionless.<br>For$\cos(Bx)$, the argument$Bx$is dimensionless:<br>$[B] [x] = [M^0 L^0 T^0]$<br>$[B] [L] = 1 \Rightarrow [B] = [L^{-1}]$<br><br>For$\sin(Dt)$, the argument$Dt$is dimensionless:<br>$[D] [t] = [M^0 L^0 T^0]$<br>$[D] [T] = 1 \Rightarrow [D] = [T^{-1}]$<br><br>Now, finding the dimensions of the required ratio$\dfrac{A D}{C B}$:<br>\left[\dfrac{A D}{C B}\right] = \left[\dfrac{A}{C}\right] \times \left[\dfrac{D}{B}\right]$
\left[\dfrac{A D}{C B}\right] = 1 \times \dfrac{[T^{-1}]}{[L^{-1}]} = [L^1 T^{-1}]$<br><br>Including mass, the dimensional formula is$[M^0 L^1 T^{-1}]$.<br><br>Answer:$[M^0 L^1 T^{-1}]