Question

Find all solutions of equation on interval [0,2pi]

1=cot^2x + cscx

Answer 1

Multiply both sides by sin^2(x).
sin^2(x)=cos^2(x)+sin(x)
sin^2(x)=1-sin^2(x)+sin(x)
2sin^2(x)-sin(x)-1=0
(2sin(x)+1)(sin(x)-1)=0
sin(x)=-1/2, x=7pi/6, x=11pi/6
or
sin(x)=1, x=pi/2

Answer 2

Thank you so much!