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**Hint:**First of all solve the quadratic equation given in the above problem that is ${{\sin }^{2}}x-2\sin x-1=0$. Reduce this quadratic equation by factorization method and then equate each factor to 0. From that, find the values of x where that sine function is vanishing. Now, it is given that in a certain interval, there are 4 solutions possible so we have to put the different values of n given in the option and see which value of n is giving 4 solutions in that interval.

**Complete step by step answer:**

The quadratic trigonometric equation given in the above problem is:

${{\sin }^{2}}x-2\sin x-1=0$

We are going to find the roots of the above equation by Shree Dharacharya rule.

Discriminant of a quadratic equation is denoted by D.

For the quadratic equation $a{{t}^{2}}+bt+c=0$ the value of D is:

$D={{b}^{2}}-4ac$

Comparing this value of D with the given quadratic equation ${{\sin }^{2}}-2\sin x-1=0$ we get,

\[\begin{align}

& D={{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -1 \right) \\

& \Rightarrow D=4+4 \\

& \Rightarrow D=8 \\

\end{align}\]

The discriminant formula for finding the roots of quadratic equation $a{{t}^{2}}+bt+c=0$ is:

$t=\dfrac{-b\pm \sqrt{D}}{2a}$

Comparing the above value of t with the given quadratic equation ${{\sin }^{2}}-2\sin x-1=0$ we get,

\[\begin{align}

& \sin x=\dfrac{2\pm \sqrt{8}}{2} \\

& \Rightarrow \sin x=\dfrac{2\pm 2\sqrt{2}}{2} \\

\end{align}\]

2 will be cancelled out from the numerator and the denominator.

\[\sin x=1\pm \sqrt{2}\]

Writing the above solution with plus sign we get,

\[\sin x=1+\sqrt{2}\]

In the above equation, we are getting the value of $\sin x$ greater than 1 which is not possible because the maximum value a sine takes is 1. Hence, this solution is rejected.

Now, writing the solution with minus sign we get,

\[\begin{align}

& \sin x=1-\sqrt{2} \\

& \Rightarrow \sin x=1-1.414 \\

& \Rightarrow \sin x=-0.414 \\

\end{align}\]

This solution is acceptable because the value is between -1 to 1. And sine can take value from -1 to 1.

Now, sine is negative in the third and fourth quadrant and from 0 to $2\pi $, sine will be negative 2 times. Hence, from 0 to $2\pi $, 2 solutions are possible and from 0 to $4\pi $, 4 different solutions are possible.

Hence, for $n=4$ in $x\in \left[ 0,n\pi \right]$ the given equation has 4 different solutions.

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

**Note:**The important concept that you extract from this problem is always check the solutions that you got from solving any quadratic equation.

For instance, in the above solution, the roots of $\sin x$ that we have got is equal to:

\[\sin x=1\pm \sqrt{2}\]

If you blindly consider both the solutions with plus and minus sign then you end up getting the wrong solution because in the above solution, we have seen when considering the solution with plus sign the solutions are non existent and we have rejected that solution.

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