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Other than the trivial solution of x = 0 , we will find the others.
Rewrite equation as:
x sin x = π 2
It is easy to see that x sin x is a strictly increasing function in ( 0 , ∞ ) and a strictly decreasing function in ( − ∞ , 0 ) . Hence x sin x = π 2 is true for two values of x , namely ( 2 π , 2 − π ) .
Therefore, the total number of solutions is: ( 0 , 2 π , 2 − π ) = 3