Find the set of possible values of tan(x+pi/6)tanx wher x is any real angle

let tan(x+pi/6)tanx=k, implies tanx(tanx+(1/√3))/(1-(tanx/√3))=k,implies tan²x +tanx((k+1/√3))-k=0, this quadratic equation in tanx has a non-negative discriminant. thus,(k+1)²/3≥ -4k, k² +14k +1≥0,the "roots" of this quadratic inequality are -7-4√3 and -7+4√3. therefore, the set of possible values is (-∞,-7-4√3] U [-7+4√3,∞)

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