Posted by **LSAT ** on Thursday, June 2, 2011 at 7:58pm.

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

1=cot^2x + cscx

- Trig -
**Mgraph**, Thursday, June 2, 2011 at 8:54pm
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

- Trig -
**Lsat**, Thursday, June 2, 2011 at 9:22pm
Thank you so much!

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