.
#cot^3x/sinx=cotx#
Let's separate the numerator of the left hand side into two pieces:
#cot^3x=cotx(cot^2x)#
#cotx(cot^2x/sinx)=cotx#
Let's divide both sides by #cotx#:
#cancelcolor(red)cotx(cot^2x/sinx)=cancelcolor(red)cotx#
#cot^2x/sinx=1#
Let's multiply both sides by #sinx#:
#sinx(cot^2x/sinx)=sinx#
#cancelcolor(red)sinx(cot^2x/cancelcolor(red)sinx)=sinx#
#cot^2x=sinx#
But we know that:
#cotx=cosx/sinx#
Let's plug this in:
#cos^2x/sin^2x=sinx#
Let's multiply both sides by #sin^2x#:
#sin^2x(cos^2x/sin^2x)=sin^3x#
#cancelcolor(red)(sin^2x)(cos^2x/cancelcolor(red)(sin^2x))=sin^3x#
#cos^2x=sin^3x#
We know that:
#sin^2x+cos^2x=1 :. cos^2x=1-sin^2x#
Let's plug this in because, by doing this, we can turn the equation into one that has all its terms in the form of the same variable:
#1-sin^2x=sin^3x#
#sin^3x+sin^2x-1=0#
This is in the form of a polynomial of degree #3#. There are various methods for solving an equation like this. You can either use a graphing utility, or rational roots theorem, or the Newton-Raphson method.
The last one is typically covered in calculus. If you have not taken calculus you can use one of the first two methods. Let's use a graphing utility in this case.
We can say:
#z=sinx# and write the equation as:
#z^3+z^2-1=0#
The graph is:
The graph crosses the #x#-axis at:
#z=0.75488#
This means:
#sinx=0.75488#
#x=arcsin(0.75488)#
#x=0.86# radians #=49.02^@#
and
#x=pi-0.86=2.28# radians #=180^@-49.02^@=130.98^@#
