What is the value of x in the equation? Thank you!

Solutions of #xtanx-1=0# are also solutions of

#xtanx=1#

which is the same equation as

#x=1/tanx = cotx#

Thus, we are looking for all values #x# such that taking the cotangent of #x# returns #x# back (i.e. all the invariant points of #cotx#).

There does not exist a closed formula for finding invariant points of a trigonometric function. However, we can find an approximation.

Using #f(x) = xtanx-1,# its derivative #f'(x) = tanx+xsec^2 x,# and the Newton-Rhapson method

#x_(n+1)=x_n-(f(x_n))/(f'(x_n))#

for finding roots of #f(x)#, we begin with an arbitrary (but useful) estimate like #x_1=pi/4~~0.7854#:

#x_2=x_1-f(x_1)/(f'(x_1))#

#color(white)(x_2)=pi/4 -f(pi/4)/(f'(pi/4))#

#color(white)(x_2)=pi/4-(pi/4tan(pi/4)-1)/(tan(pi/4)+pi/4 sec^2 (pi/4))#

#color(white)(x_2)=pi/4-(pi/4-1)/(1+pi/4(2))#

#color(white)(x_2)=pi/4(1-(2(pi-4))/(pi(2+pi)))" " ~~ 0.868875#

A second iteration of the Newton-Rhapson method produces

#x_3=x_2-f(x_2)/(f'(x_2))#

#color(white)(x_3)~~0.868875-f(0.868875)/(f'(0.868875))#

#color(white)(x_3)~~0.868875-0.027551/(3.266703)#

#color(white)(x_3)~~0.860441#

A third produces

#x_4=x_3-f(x_3)/(f'(x_3))" "~~0.860333#

A fourth:

#x_5~~0.860334#

Our estimate of the (first positive) root of #f(x)# is now accurate to a precision of at least 5 decimal digits. Repeat the iteration procedure until the desired precision is reached.