If $$\frac{3+i\sin\theta}{4-i\cos\theta}$$, $$\theta \in [0, 2\pi]$$, is a real number, then an argument of $$\sin\theta + i\cos\theta$$ is

We have the complex expression $$z=\dfrac{3+i\sin\theta}{4-i\cos\theta},\qquad\theta\in[0,\,2\pi].$$

Because $$z$$ is given to be a real number, its imaginary part must vanish. A convenient way to isolate the imaginary part is to multiply the numerator and the denominator by the conjugate of the denominator, making the new denominator purely real.

The conjugate of $$\,4-i\cos\theta\,$$ is $$\,4+i\cos\theta.$$ Multiplying, we get

$$$z=\dfrac{3+i\sin\theta}{4-i\cos\theta}\cdot\dfrac{4+i\cos\theta}{4+i\cos\theta} =\dfrac{(3+i\sin\theta)(4+i\cos\theta)} {(4-i\cos\theta)(4+i\cos\theta)}.$$$

In the denominator we use the identity $$(a-b)(a+b)=a^{2}-b^{2}$$ with $$a=4$$ and $$b=i\cos\theta$$:

$$$(4-i\cos\theta)(4+i\cos\theta)=4^{2}-(i\cos\theta)^{2}=16-(-\cos^{2}\theta)=16+\cos^{2}\theta.$$$

Thus the denominator is the real number $$16+\cos^{2}\theta.$$ Now we concentrate on the numerator

$$(3+i\sin\theta)(4+i\cos\theta).$$

We expand it term by term:

$$$\begin{aligned} (3+i\sin\theta)(4+i\cos\theta) &=3\cdot4+3\cdot(i\cos\theta)+i\sin\theta\cdot4+i\sin\theta\cdot(i\cos\theta)\\ &=12+3i\cos\theta+4i\sin\theta+i^{2}\sin\theta\cos\theta\\ &=12+3i\cos\theta+4i\sin\theta-\sin\theta\cos\theta\\ &=(12-\sin\theta\cos\theta)+i\bigl(3\cos\theta+4\sin\theta\bigr). \end{aligned}$$$

The denominator is real, so for $$z$$ itself to be real, the imaginary part of the numerator must be zero:

$$3\cos\theta+4\sin\theta=0.$$

Dividing both sides by $$\cos\theta$$ (which is permissible wherever $$\cos\theta\neq0$$), we obtain

$$3+4\tan\theta=0\quad\Longrightarrow\quad\tan\theta=-\frac{3}{4}.$$

The acute angle whose tangent is $$\dfrac{3}{4}$$ is $$\alpha=\tan^{-1}\!\left(\frac{3}{4}\right).$$ Since the tangent is negative, $$\theta$$ must lie in the second or fourth quadrants. Hence

$$$\theta=\pi-\alpha\quad\text{or}\quad\theta=2\pi-\alpha,\qquad \text{where } \alpha=\tan^{-1}\!\left(\frac{3}{4}\right).$$$

Now we must find an argument of the complex number $$w=\sin\theta+i\cos\theta.$$

First observe the trigonometric identities

$$$\cos\!\left(\frac{\pi}{2}-\theta\right)=\sin\theta,\qquad \sin\!\left(\frac{\pi}{2}-\theta\right)=\cos\theta.$$$

Therefore

$$$w=\sin\theta+i\cos\theta =\cos\!\left(\frac{\pi}{2}-\theta\right)+i\sin\!\left(\frac{\pi}{2}-\theta\right) =e^{\,i\left(\frac{\pi}{2}-\theta\right)}.$$$

Thus one argument of $$w$$ is

$$\arg w=\frac{\pi}{2}-\theta.$$ Substituting each permissible $$\theta$$:

For $$\theta=\pi-\alpha$$ we get

$$$\arg w=\frac{\pi}{2}-(\pi-\alpha)=\frac{\pi}{2}-\pi+\alpha =-\frac{\pi}{2}+\alpha.$$$

For $$\theta=2\pi-\alpha$$ we get

$$$\arg w=\frac{\pi}{2}-(2\pi-\alpha)=\frac{\pi}{2}-2\pi+\alpha =-\frac{3\pi}{2}+\alpha.$$$

The two expressions differ by exactly $$-\,\pi$$, so they represent the same set of arguments (they vary by an integral multiple of $$\,\pi$$). To compare with the options, it is convenient to add $$\,\pi$$ to the first value:

$$-\frac{\pi}{2}+\alpha+\pi=\frac{\pi}{2}+\alpha =\pi-\bigl(\tfrac{\pi}{2}-\alpha\bigr).$$

But $$\tfrac{\pi}{2}-\alpha=\tan^{-1}\!\left(\frac{4}{3}\right)$$ because for complementary acute angles we have $$\tan\beta=\frac{1}{\tan\alpha}$$. Hence

$$\arg w=\pi-\tan^{-1}\!\left(\frac{4}{3}\right).$$

This value appears exactly as Option A.

Hence, the correct answer is Option A.