Tan(2x)-cot(x)=0?

Question

I solved it and got the solutions: pi/6, 5pi/6, 7pi/6 and 11pi/6. Why did I not get the pi/2 and 3pi/2 solutions?

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Answer 1 · 2018-02-18T02:01:12

My solution is found in explanation section.

$tan2x-cotx=0$

$tan2x=cotx$

$tan2x*tanx=1$

$(sin2x*sinx)/(cos2x*cosx)=1$

$[1/2*(cosx-cos3x)]/[1/2*(cos3+cosx)]=1$

$(cosx-cos3x)/(cos3x+cosx)=1$

$cosx-cos3x=cos3x+cosx$

$2cos3x=0$

$cos3x=0$

I have $2$ states for $cos3x=0$ ,

$a)$ $cos3x=cos(pi/2+2pi*k)$

$3x=pi/2+2pi*k$ , so $x=pi/6+(2pi)/3*k$

Consequently, $x_1=pi/6$ , $x_2=(5pi)/6$ and $x_3=(3pi)/2$ for $k=0, 1$ and $2$

$b)$ $cos3x=cos((3pi)/2+2pi*k)$

$3x=(3pi)/2+2pi*k$ , so $x=pi/2+(2pi)/3*k$

Consequently, $x_4=pi/2$ , $x_5=(7pi)/6$ and $x_6=(11pi)/6$ for $k=0, 1$ and $2$