Question 907463
Use an algebraic method to find the solution for 0<=x<=2pi of the equation 
5cot x + 2cosec^2 x = 5
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{{{5cot(x)+2csc^2(x)=5}}}
csc^2(x)=cot(x)+1
{{{5cot(x)+2(cot^2(x)+1)=5}}}
{{{5cot(x)+2cot^2(x)+2=5}}}
{{{2cot^2(x)+5cot(x)-3=0}}}
(cot(x)+3)(2cot(x)-1)=0
cot(x)=-3
(Using inverse tan key of calculator set to radians)
x&#8776;2.82, 5.96 (In quadrants II and IV in which cot<0)
or
cot(x)=1/2
x&#8776;1.11, 4.25 (In quadrants I and III in which cot>0)