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draw a graph..
it comes out to be the same graph as that of
|sin x|
now
x+|x|, for x>0, it is y=2x
for x<0, y=0
at every n(pi), value of f(x)=|sin x|=0
so -pi, -2 pi and 0 are solutions
for y=2x and y= |sin x|
sin x <=x for x>0
so sin x< 2x
so no solution in first quadrant
so only 3 solns
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Moderation Team
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