The maximum value of $$3\cos\theta + 5\sin\left(\theta - \frac{\pi}{6}\right)$$ for any real value of $$\theta$$ is:

We have to find the greatest value that the expression

$$E(\theta)=3\cos\theta+5\sin\left(\theta-\frac{\pi}{6}\right)$$

can take for any real $$\theta$$. To compare unlike trigonometric functions, we first express everything in terms of the basic $$\sin\theta$$ and $$\cos\theta$$ functions.

We recall the compound-angle identity for sine:

$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.$$

Using this with $$\alpha=\theta$$ and $$\beta=\dfrac{\pi}{6}$$, we write

$$\sin\left(\theta-\frac{\pi}{6}\right)=\sin\theta\cos\frac{\pi}{6}-\cos\theta\sin\frac{\pi}{6}.$$

Now we substitute the exact values $$\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$$ and $$\sin\dfrac{\pi}{6}=\dfrac12$$:

$$\sin\left(\theta-\frac{\pi}{6}\right)=\sin\theta\left(\frac{\sqrt3}{2}\right)-\cos\theta\left(\frac12\right).$$

Next, we insert this expansion into the original expression:

$$E(\theta)=3\cos\theta+5\Bigg[\sin\theta\left(\frac{\sqrt3}{2}\right)-\cos\theta\left(\frac12\right)\Bigg].$$

Multiplying the 5 inside the brackets term-by-term gives

$$E(\theta)=3\cos\theta+\frac{5\sqrt3}{2}\sin\theta-\frac52\cos\theta.$$

Now we collect the like $$\cos\theta$$ terms together:

$$3\cos\theta-\frac52\cos\theta=\left(3-\frac52\right)\cos\theta=\frac12\cos\theta.$$

Hence the entire expression simplifies to

$$E(\theta)=\frac12\cos\theta+\frac{5\sqrt3}{2}\sin\theta.$$

For clarity, we rename the coefficients:

Let $$A=\frac{5\sqrt3}{2} \quad\text{and}\quad B=\frac12.$$

Thus we have

$$E(\theta)=A\sin\theta+B\cos\theta.$$

There is a standard result for an expression of the form

$$A\sin\theta+B\cos\theta,$$

namely:

It can always be rewritten as

$$\sqrt{A^{2}+B^{2}}\;\sin(\theta+\phi)$$

for some phase angle $$\phi$$ (because $$\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi$$). Since the sine function itself is bounded between $$-1$$ and $$1$$, the maximum possible value is precisely the amplitude, i.e.

$$\max\bigl(A\sin\theta+B\cos\theta\bigr)=\sqrt{A^{2}+B^{2}}.$$

Therefore we compute

$$A^{2}+B^{2}=\left(\frac{5\sqrt3}{2}\right)^{2}+\left(\frac12\right)^{2} =\frac{25\cdot3}{4}+\frac{1}{4} =\frac{75+1}{4} =\frac{76}{4} =19.$$

Taking the square root gives

$$\sqrt{A^{2}+B^{2}}=\sqrt{19}.$$

This is the largest value that $$E(\theta)$$ can ever reach for any real $$\theta$$.

Hence, the correct answer is Option A.