Question 885200
How do you solve
3cot^2-3cotx=1
for interval: [0,π]
3cotx-3cotx-1=0
solve for cotx using quadratic formula:
 {{{cotx = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=3, b=-3, c=-1
ans:
cotx≈-0.26
tanx=1/cotx=1/-0.26=-3.85
x=1.82 radians
or
cotx≈1.26
tanx=1/cotx=1/1.26=0.79
x=0.67 radians

ans rounded to 2 decimal places