let

3sin6t+4cos6t=k sin (6t + α) =a sin 6t cos α + a cos 6t sin α

> a cos α = 3 ... ( 1 ) and a sin α = 4 ... ( 2 )

Squarring and adding eqns. ( 1 ) and ( 2 ),

=> a^2 = 25 => a = 5

=> cos α = 3/5 and sin α = 4/5

Thus, 3 sin 6t - 4 cos 6t = 5 sin (6t + α),

where, value of α is obtained from cos α=3/5 and sin α=4/5

This method of simplification helps to find the range of the function 3 sin 6t - 4 cos 6t as the range of 5 sin (6t + α) which is ± 5.

Thus, the range of the function asin x - bcos x is

± √[(a)^2 + (b)^2].