Solve the equation $2cos^2xcotx=cotx$ within the interval $0<=x<2pi$ ?

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Answer 1 · 2017-03-25T16:28:57

$x=npi+pi/2$ or $x=npi+-pi/4$ and within $0<=x<2pi$ , we have $x={pi/4,pi/2,(3pi)/4,(5pi)/4,(3pi)/2,(7pi)/4}$

$2cos^2xcotx=cotx$

$hArr2cos^2xcotx-cotx=0$

or $cotx(2cos^2x-1)=0$

or $cotx(sqrt2cosx-1)(sqrt2cosx+1)=0$

Hence either $cotx=0$ i.e. $x=npi+pi/2$

or $cosx=1/sqrt2=cospi/4$ or $cosx=-1/sqrt2=cos((3pi)/4)$

Hence $x=npi+-pi/4$

and within $0<=x<2pi$ , we have $x={pi/4,pi/2,(3pi)/4,(5pi)/4,(3pi)/2,(7pi)/4}$

Solve the equation 2cos^2xcotx=cotx2cos^2xcotx=cotx within the interval 0<=x<2pi0<=x<2pi?