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!