Question 534177: find all the solutions on the interval [0,2pi)
2cot^2x+3cscx=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find all the solutions on the interval [0,2pi)
2cot^2x+3cscx=0
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2(cos^2x/sin^2x)+3(1/sinx)=0
(2cos^2x/sin^2x)+(3/sinx)=0
LCD: sin^2x
(2cos^2x)+(3sinx)=0
2(1-sin^2x)+3sinx=0
2-2sin^2x+3sinx=0
2sin^2x-3sinx-2=0
(2sinx+1)(sinx-2)=0
..
2sinx=1=0
sinx=-1/2
x=arcsin(-1/2)=7π/6 and 11π/6 (in quadrants III and IV where sin<0)
..
sinx-2=0
sinx≠2 (reject, (-1≤sinx≤1)
solutions:7π/6 and 11π/6
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