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(C) Given equation is $a^2 - 2a\sin x + 1 = 0$. For $a$ to be a real value,the discriminant $D$ of this quadratic equation in $a$ must be greater than

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$2$

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(C) Given equation is $a^2 - 2a\sin x + 1 = 0$. For $a$ to be a real value,the discriminant $D$ of this quadratic equation in $a$ must be greater than or equal to $0$. $D = (-2\sin x)^2 - 4(1)(1) = 4\sin^2 x - 4 = 4(\sin^2 x - 1) = -4\cos^2 x$. Since $D \ge 0$,we must have $-4\cos^2 x \ge 0$,which implies $\cos^2 x \le 0$. Since $\cos^2 x$ cannot be negative,the only possibility is $\cos^2 x = 0$,which means $\cos x = 0$. If $\cos x = 0$,then $\sin x = 1$ or $\sin x = -1$. Case $1$: If $\sin x = 1$,the equation becomes $a^2 - 2a + 1 = 0$,which is $(a - 1)^2 = 0$,giving $a = 1$. Case $2$: If $\sin x = -1$,the equation becomes $a^2 + 2a + 1 = 0$,which is $(a + 1)^2 = 0$,giving $a = -1$. Thus,the real values of $a$ are $1$ and $-1$. There are $2$ such real values.

The number of real values of $a$ satisfying the equation $a^2 - 2a\sin x + 1 = 0$ is

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