The equation \[a\sin x + b\cos x = c\], where \[\left| c \right| > \sqrt {{a^2} + {b^2}} \] has
1) A unique solution
2) Infinite number of solutions
3) No solution
4) None of the above

Hint: We will first consider the given equation and as we have to determine the category of the solution so, we will start by finding the value of \[c\]. Now compare this value with the given value that is, \[\left| c \right| > \sqrt {{a^2} + {b^2}} \] and determine whether the solution has no solution or infinite solution or unique solution.

Complete step by step solution:
Consider the given equation, \[a\sin x + b\cos x = c\].
Here, we have to check that the given equation has a unique solution or the infinite number of solutions or no solution.
We know that the maximum value of \[a\sin x + b\cos x\] is \[\sqrt {{a^2} + {b^2}} \] and minimum value of \[a\sin x + b\cos x\] is \[ - \sqrt {{a^2} + {b^2}} \].
So, we can say that
\[
   - \sqrt {{a^2} + {b^2}} < a\sin x + b\cos x < \sqrt {{a^2} + {b^2}} \\
  \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left| {a\sin x + b\cos x} \right| \leqslant \sqrt {{a^2} + {b^2}} \\
 \]

In this question \[a\sin x + b\cos x = c\].
So, we have \[\left| {a\sin x + b\cos x} \right| = \left| c \right|\]
Therefore, \[\left| c \right| \leqslant \sqrt {{a^2} + {b^2}} \].
But according to the given condition in question, \[\left| c \right| > \sqrt {{a^2} + {b^2}} \] which is different from what we have obtained above.
Hence, \[\left| c \right| \leqslant \sqrt {{a^2} + {b^2}} \] and \[\left| c \right| > \sqrt {{a^2} + {b^2}} \] are two different scenarios.
And both of these conditions cannot be true simultaneously. So, we can say that the equation \[a\sin x + b\cos x = c\], where \[\left| c \right| > \sqrt {{a^2} + {b^2}} \] does not have a solution.
Therefore, the correct option is C.

Note: In this question, first of all, note that the range of \[c\] is given. We can find another range of \[c\] by evaluating the value of \[c\] using the given equation. If the calculated range is same then we can say that there is a solution. If that calculated range is different from the given range, then it will be a contradiction to our hypothesis. Hence, there will be no solution. Also, the maximum value of \[a\sin x + b\cos x\] is \[\sqrt {{a^2} + {b^2}} \] and minimum value of \[a\sin x + b\cos x\] is \[ - \sqrt {{a^2} + {b^2}} \].