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2015-06-18T21:37:41+05:30

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           Tan x =  x - x³
           Let Y = tan x - x + x³
           we want solutions of  Y = Tan x - x + x³ = 0

We can see that  at x = 0 ,  the equation is satisfied as Tan x = x - x³.
      So x = 0 is one solution.

Derivative of Tan x = Sec² x  > 0 for  0 <= x <= 1.
   So Tan x is rising continuously in the interval  0<= x <= 1.

Derivative of x - x³ = 1 - 3 x²  > 0  for  0<= x < 1/√3.
                                           < 0  for  1/√3 < x <= 1
           so during the interval 1/√3 < x < 1,  x - x³ decreases.

At   x = 1/√3,  Tan x = 0.6518
                     x - x³ =  0.3849

Thus we find that value of Tan x >    x - x³    for  0< x <= 1

we apply similar principle for the 3rd quadrant too.
      so,    Tan x  <  x - x³   for  -1 <= x < 0.  

Hence, there is only one solution at x = 0.
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Derivatives of  tan x  and  x - x³  we find.
       d tan x  / dx = sec² x    > 1      we know that secant is always > 1.
       So y = tan x curve is above  y = x.

     y =  x - x³
     slope of the curve =  d y / dx = 1 - 3 x²    < 1 

Hence,   tan x always rises faster than  x - x³  for x > 0  or  falls faster than x - x³ for x < 0.  Hence they meet only at  x = 0.

There is only one solution for the above equation.

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