Splitting a trigonometric equation into casesI am not understanding this step in a solution I found. We are solving y = − 2 y 1 − y 2 where y = t a n ( x )The solution says that we can now split this into two cases, case 1 where y = 0 and case 2 where y ≠ 0, I do not understand how we know that we can split up into two cases at this point, please advise. What's the motivation?(As a side note, the cases arecase 1, y=0 x = k π , k ∈ Z case 2, y ≠ 0 1 − y 2 = − 2 ⟺ y 2 = 3then x = ± π 3 + k π , k ∈ Z ).

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Splitting a trigonometric equation into casesI am not understanding this step in a solution I found. We are solving  y  =            −      2      y              1      −              y        2            where   y  =  t  a  n  (  x  )The solution says that we can now split this into two cases, case 1 where   y  =  0 and case 2 where   y  ≠  0, I do not understand how we know that we can split up into two cases at this point, please advise. What&#039;s the motivation?(As a side note, the cases arecase 1, y=0  x  =  k  π  ,  k  ∈      Z  case 2,   y  ≠  0  1  −      y    2    =  −  2    ⟺        y    2    =  3then  x  =  ±      π    3    +  k  π  ,  k  ∈      Z  ).

Travis Sellers · Accepted Answer

One would be tempted to cancel out the term   y from the numerators on both sides. i.e. write it as  1  −      y    2    =            −      2      y        y    =  −  2. That would be valid only when   y  ≠  0, since you would be dividing   0 by   0 when   y  =  0. Hence the motivation to treat   y  =  0 as a special case.

Splitting a trigonometric equation into cases I am not understanding this step in a solution I found. We are solving y=(-2y)/(1-y^2) where y=tan(x).

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