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Does #x^3+sin x = 1# have a root in #(0, 2pi)# ?

Answer 1

Daniel Acosta

Yes, it has one root in #(0, 2pi)#

Let #f(x) = x^3+sin x - 1#, so zeros of #f(x)# are roots of our equation.

Note that:

#f(0) = 0+0-1 = -1 < 0#

#f(pi/2) = pi^3/8+1-1 = pi^3/8 > 0#

Both #x^3# and #sin x# are continuous and strictly monotonically increasing in #(0, pi/2]#, so #f(x)# is too.

#pi/2 ~~ 1.57 > root(3)(2)#, so when #x >= pi/2#, #f(x) > 2+sin(x)-1 > 0#

Hence by Bolzano's theorem #f(x)# has exactly one zero in #(0, 2pi)#, actually somewhere in #(0, pi/2)#.

graph{x^3+sinx -1 [-7, 7, -20, 20]}

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