#### Answer

$$h\left( t \right) = - 2\cos 3t + 6$$

#### Work Step by Step

$$\eqalign{
& h'\left( t \right) = 6\sin 3t \cr
& h\left( t \right) = \int {h'\left( t \right)} dt \cr
& then \cr
& h\left( t \right) = \int {6\sin 3t} dt \cr
& find{\text{ the general solution}} \cr
& h\left( t \right) = - 2\cos 3t + C \cr
& {\text{using the initial condition }}h\left( {\pi /6} \right) = 6 \cr
& 6 = - 2\cos 3\left( {\frac{\pi }{6}} \right) + C \cr
& 6 = - 2\cos \left( {\frac{\pi }{2}} \right) + C \cr
& C = 6 \cr
& {\text{the solution to the initial value problem is}} \cr
& h\left( t \right) = - 2\cos 3t + 6 \cr} $$