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**Hint:**Using the equation ${\sin ^2}x + {\cos ^2}x = 1$ change the equation into a quadratic equation in sine or cosine function. Now find the discriminant of the quadratic equation to find out the characteristics of the solution of the equation. Since a sine or cosine function can have imaginary value, so the discriminant must be non-negative for real solutions.

**Complete step-by-step answer:**

We know that: ${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x} $

Using the above relationship, we can express $\cos x$ in the form of $\sin x$. So, now our given equation will become:

$ \Rightarrow \sqrt 3 \sin x + \sqrt {1 - {{\sin }^2}x} = 4$

$ \Rightarrow \sqrt {1 - {{\sin }^2}x} = 4 - \sqrt 3 \sin x$

Now squaring both the sides of the above equation, we have a simpler equation

$ \Rightarrow 1 - {\sin ^2}x = {\left( {4 - \sqrt 3 \sin x} \right)^2}$

Let us now expand this by using ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and simplifying it further

$ \Rightarrow 1 - {\sin ^2}x = {4^2} + {\left( {\sqrt 3 \sin x} \right)^2} - 2 \times 4 \times \sqrt 3 \sin x$

$ \Rightarrow 1 - {\sin ^2}x = 16 + 3{\sin ^2}x - 8\sqrt 3 \sin x$

$ \Rightarrow 4{\sin ^2}x - 8\sqrt 3 \sin x + 15 = 0………...(i)$

Here we got a quadratic equation in $\sin x$ and we can check for the roots for this equation.

It is important to understand a few things about quadratic equations first. If we have an equation of the form $a{x^2} + bx + c = 0$. We can use the expression,${b^2} - 4ac$, also known as the discriminant to determine the number of roots of solutions in a quadratic equation. There are three cases:

${b^2} - 4ac < 0$: The equation has zero real solutions. The graph does not cross the x-axis.

${b^2} - 4ac = 0$: The equation has one real solution. The graph touches the x-axis at one point.

${b^2} - 4ac > 0$: The equation has two real solutions. The graph crosses through the x-axis at two points.

Now let us use this same concept for solving this problem.

For our previous equation $(i)$, ${b^2} - 4ac = {\left( {8\sqrt 3 } \right)^2} - 4 \times 4 \times 15 = 64 \times 3 - 16 \times 15 = 192 - 240 = - 48 < 0$

So here the discriminant is negative, which signifies that this quadratic equation has no real solutions, which means there is no real value of $\sin x$ to satisfy this equation.

Therefore, we can conclude that the equation $\sqrt 3 \sin x + \cos x = 4$ does not have any solution.

**So, the correct answer is “Option B”.**

**Note:**An alternative approach can be taken in changing the given equation into a quadratic equation in cosine function. You can also approach it by changing the given equation using \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\] or \[\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b\] .

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