# Please I need help in these exercises: tan (x+π/4)=2tanx + 2

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Solve `tan(x+pi/4)=2tanx+2` :

Now `tan(A+B)=(tanA+tanB)/(1-tanAtanB)` so we can rewrite the left-hand side:

`(tanx+tan(pi/4))/(1-tanxtan(pi/4))=2tanx+2` and `tan(pi/4)=1` so we have:

`(tanx+1)/(1-tanx)=2tanx+2` Multiply both sides by (1-tanx)

`tanx+1=2-2tan^2x`

`2tan^2x+tanx-1=0`

`(2tanx-1)(tanx+1)=0` By the zero product property

`tanx=1/2 "or" tanx=-1`

If tanx=-1 then `x=-pi/4+npi,n in ZZ` (n an integer)

If `tanx=1/2 ==> x=tan^(-1)1/2==>x~~.464+npi,n in ZZ`

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The solutions are `x=-pi/4+npi,x~~.464+npi`

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The graph of the left side in black, the right side in red: