The number of integral values of $$k$$ for which the equation $$3\sin x + 4\cos x = k + 1$$ has a solution, $$k \in R$$ is ______.

The equation is $$3\sin x + 4\cos x = k + 1$$.

The expression $$3\sin x + 4\cos x$$ has the form $$a\sin x + b\cos x$$, whose range is $$[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$$.

Here $$\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$.

So $$3\sin x + 4\cos x \in [-5, 5]$$, and for the equation to have a solution we need $$-5 \leq k + 1 \leq 5$$, which gives $$-6 \leq k \leq 4$$.

The integral values of $$k$$ in this range are $$k = -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4$$.

The number of integral values is $$4 - (-6) + 1 = 11$$.