# Find x where 0<=x<=2pi sin x - cot^2 x = 1 Anyone know how to do these things??

##### 1 Answer

May 21, 2018

We can rewrite:

#sinx = 1 + cot^2x#

Now recall that

#sinx = csc^2x#

#sinx -1/sin^2x = 0#

#(sin^3x - 1)/sin^2x= 0#

#sin^3x =1 #

#sinx = 1#

#x = pi/2#

We can confirm graphically our answer is correct.

Hopefully this helps!