Consider the system of equations $ \sin x\cos 2y={{\left( {{a}^{2}}-1 \right)}^{2}}+1,\cos x\sin 2y=a+1 $ . The number of values of $ a $ for which the system has a solution is \[\]
A. $ 1 $ \[\]
B. $ 2 $ \[\]
C. $ 3 $ \[\]
D. infinite \[\]

Hint: We recall the ranges of sine and cosine trigonometric function as $ \left[ -1,1 \right] $ . We use the product of the ranges to deduce that the product of sine and cosine function will also in the interval $ \left[ -1,1 \right] $ . We use these restrictions to find values of $ a $ and see if that satisfies the system of equations .\[\]

Complete step by step answer:
We know that both sine and cosine function have the real number set as their domain and the closed interval $ \left[ -1,1 \right] $ as their range. It means the minimum of value of sine and cosine function is $ -1 $ and the maximum value is $ 1 $ . Hence the product of sine and cosine will also lie between $ -1 $ and 1 because $ -1\le ab\le 1 $ for all $ a,b $ such that $ \left| a \right|\le 1,\left| b \right|\le 1 $ .
We are given the system of two equations as follows;
\[\begin{align}
  & \sin x\cos 2y={{\left( {{a}^{2}}-1 \right)}^{2}}+1....\left( 1 \right) \\
 & \cos x\sin 2y=a+1....\left( 2 \right) \\
\end{align}\]
Since the product of sine and cosine will also lie between $ -1 $ and 1 we have;
\[\begin{align}
  & -1\le {{\left( {{a}^{2}}-1 \right)}^{2}}+1\le 1\text{ }......\left( 3 \right) \\
 & -1\le a+1\le 1........\left( 4 \right) \\
\end{align}\]
Let us observe the first condition since a square is always non negative we have $ \left( {{a}^{2}}-1 \right)\ge 0 $ . Then we have;
\[{{\left( {{a}^{2}}-1 \right)}^{2}}+1\ge 1....\left( 5 \right)\]
 We have from (3) that
\[{{\left( {{a}^{2}}-1 \right)}^{2}}+1\le 1...\left( 6 \right)\]
So we have from inequalities (5) and (6) that
\[\begin{align}
  & {{\left( {{a}^{2}}-1 \right)}^{2}}+1=1 \\
 & \Rightarrow {{\left( {{a}^{2}}-1 \right)}^{2}}=0 \\
 & \Rightarrow {{a}^{2}}-1=0 \\
 & \Rightarrow a=-1,1 \\
\end{align}\]
If we take $ a=1 $ the inequality (4) is not satisfied since $ a+1\le 1 $ . So the only possible value of $ a $ which satisfies the inequalities and hence valid value for the system of equation is $ a=-1 $ . So the correct choice is A. \[\]

Note:
We note that if we put $ a=1 $ in the given system of equations we shall obtain the equations $ \sin x\cos 2y=1,\cos x\sin 2y=0 $ . We can add the respective sides of the equation a and the sine compound angle formula $ \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ to have the equation $ \sin \left( 2y+x \right)=1 $ whose solution is $ 2y+x=\left( 2n+1 \right)\dfrac{\pi }{2} $ where $ n $ is any integer. We can directly use the sine compound angle formula to find possible values of $ a $ .