To do problems like this, one way to do that would be rewrite the expression in the form of y = A sin (bx + c), so that you can immediately identify the period which would be \frac{2 \pi}{b} and the amplitude which would be A.

We will repeatedly use the formula \sin (A + B) = \sin A \cos B + \cos A \sin B to achieve this.

Let's first expand \sin (x - 3) = \sin x \cos 3 + \cos x \sin 3 .

Then f(x) = (3 + 4 \cos 3) \sin x + 4 \sin 3 \cos x .

Then f(x) = A \sin (x + B) where A = \sqrt{(3 + 4 \cos 3)^2 + 4 \sin^2 3} = \sqrt{9 + 4 \cos^2 3 + 24 \cos 3 + 4 \sin^2 3} = \sqrt{13 + 24 \cos 3} and B=\sin^{-1}\left(\frac{4\sin3}{A}\right).

Therefore, the period is 2 \pi and the amplitude is \sqrt{13 + 24 \cos 3} .

I hope this helps.