Here, $${y_1} = 4\sin (2x - 6t)$$

$${y_2} = 3\sin \left( {2x - 6t - {\pi \over 2}} \right)$$

The phase difference between two waves is $$\phi = {\pi \over 2}$$

The amplitude of the resultant wave is

$$A = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } $$

$$ = \sqrt {{4^2} + {3^2} + 2 \times 4 \times 3 \times \cos {\pi \over 2}} = 5$$