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Solve the equation, (1-tanx)(1+sin2x)=1+...

Solve the equation, `(1-tanx)(1+sin2x)=1+tanx`

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`(1-tanx) (1+(2tanx)/(1+tan^(2)x))=1+tanx`
`rArr (1-tanx)(1+tan^(2)x+2tanx)/(1+tan^(2)x)=1+tanx`
`rArr (1-tanx)(1+tanx)^(2) -(1+tanx)(1+tan^(2)x)=0`
`rArr (1+tanx)(1-tan^(2)x-1-tan^(2)x)=0`
which breaks into
`tanx =0 rArr x = npi, n int Z`
and `1+tanx=0 rArr tanx =-1= tan(-pi/4) therefore x=npi -pi/4, n int Z`
Then the solutions are
`x=npi, n int Z`
`x=npi - pi/4, n int Z`
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