The number of real roots of the equationtan x- 6-2 tan x- for x -0-3 - are

The correct option is A nbsp;are nbsp;given nbsp;thattanx+π6=2 nbsp;tanx⇒tanx+tanπ61 tanx nbsp;tanπ6=2 nbsp;tanxLet nbsp;t=tanx, nbsp;we nbsp ...

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Standard XII

Mathematics

General Solution of tan theta = tan alpha

The number of...

Question

$The number of real roots of the equation tan (x + \frac{π}{6}) = 2 tan x, for x \in (0, 3 π) are$
.

0.0

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Solution

The correct option is A 0.0
$We are given that tan (x + \frac{π}{6}) = 2 tanx \Rightarrow \frac{tanx + tan \frac{π}{6}}{1 - tanx tan \frac{π}{6}} = 2 tanx Let t = tanx, we get \Rightarrow \frac{t + \frac{1}{\sqrt{3}}}{1 - \frac{t}{\sqrt{3}}} = 2 t \Rightarrow t + \frac{1}{\sqrt{3}} = 2 t (1 - \frac{t}{\sqrt{3}}) \Rightarrow \sqrt{3} t + 1 = 2 \sqrt{3} t (1 - \frac{t}{\sqrt{3}}) \Rightarrow \sqrt{3} t + 1 = 2 \sqrt{3} t - 2 t^{2} \Rightarrow 2 t^{2} - \sqrt{3} t + 1 = 0 For the above quadratic equation Discriminant D = 3 - 8 < 0, Hence, the equation has no real solution .$

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The number of real roots of the equationtan(x+π6)=2 tanx, for x∈(0, 3π) are .

$The number of real roots of the equation tan (x + \frac{π}{6}) = 2 tan x, for x \in (0, 3 π) are$
.